Okay, so I need to find the real roots of this equation:

\[
\frac{(x+1)(x-3)}{5(x+2)(x-4)} + \frac{(x+3)(x-5)}{9(x+4)(x-6)} - \frac{2(x+5)(x-7)}{13(x+6)(x-8)} = \frac{92}{585}
\]

Hmm, this looks pretty complicated. Let me start by understanding each part of the equation. Each term is a rational function, and the equation is set equal to 92/585. The denominators are all products of linear terms, so I need to be careful about the values of x that would make the denominators zero. Those would be x = -2, 4, -4, 6, -6, 8. So x cannot be any of those values. Otherwise, the function is defined.

Now, the equation has three fractions with different denominators. To solve this, maybe I can combine them into a single fraction by finding a common denominator. The denominators are 5(x+2)(x-4), 9(x+4)(x-6), and 13(x+6)(x-8). The common denominator would be the product of all distinct linear factors in the denominators. Let's list them:

From the first term: 5, (x+2), (x-4)
Second term: 9, (x+4), (x-6)
Third term: 13, (x+6), (x-8)

So the common denominator would be 5*9*13*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8). That's a huge denominator! But maybe if I multiply both sides by this common denominator, the equation will simplify.

Let me write that down. Multiply both sides by 5*9*13*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8). That would eliminate all denominators. Let me compute the left-hand side terms:

First term: [ (x+1)(x-3) ] / [5(x+2)(x-4) ] multiplied by the common denominator becomes [ (x+1)(x-3) ] * 9*13*(x+4)(x-6)(x+6)(x-8)

Similarly, the second term: [ (x+3)(x-5) ] / [9(x+4)(x-6) ] multiplied by the common denominator becomes [ (x+3)(x-5) ] * 5*13*(x+2)(x-4)(x+6)(x-8)

Third term: [ 2(x+5)(x-7) ] / [13(x+6)(x-8) ] multiplied by the common denominator becomes [ 2(x+5)(x-7) ] * 5*9*(x+2)(x-4)(x+4)(x-6)

And the right-hand side: 92/585 multiplied by the common denominator is 92/585 * 5*9*13*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8)

Wait, but 5*9*13 is 585, so 92/585 * 585 is 92. Therefore, the right-hand side simplifies to 92*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8)

So putting it all together, the equation becomes:

[ (x+1)(x-3)*9*13*(x+4)(x-6)(x+6)(x-8) ] + [ (x+3)(x-5)*5*13*(x+2)(x-4)(x+6)(x-8) ] - [ 2(x+5)(x-7)*5*9*(x+2)(x-4)(x+4)(x-6) ] = 92*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8)

Wow, this seems really messy. Expanding all of these terms would be tedious. Maybe there's a smarter way. Let me see if there's a pattern or a substitution that can simplify this.

Looking at the denominators again, each term has a denominator that is a product of two linear factors, each separated by 2. For example, (x+2)(x-4) = (x+2)(x - (2+2)), (x+4)(x-6) = (x+4)(x - (4+2)), etc. So maybe there's some symmetry here.

Alternatively, perhaps if I substitute y = x - something, but I don't see an obvious substitution. Alternatively, maybe note that the denominators are similar but shifted. Let me check.

Wait, the numerators also have similar factors. For example, the first numerator is (x+1)(x-3), which is (x+1)(x - (3)). The second numerator is (x+3)(x-5), which is (x+3)(x -5). The third numerator is 2(x+5)(x-7). So each numerator is increasing by 2 in the constants. Similarly, the denominators for each term have factors that are also increasing by 2. For example, the first denominator is (x+2)(x-4), the second is (x+4)(x-6), the third is (x+6)(x-8). So the denominators are also shifted by 2 each time. Maybe this pattern can be exploited.

Alternatively, maybe if I consider the entire equation as a function f(x) = left-hand side - right-hand side = 0, then f(x) is equal to zero. Since the equation is a rational function, the roots would be the solutions where the numerator equals zero after combining all terms. But expanding the numerator would be too tedious.

Alternatively, maybe test some integer values of x that are not excluded (i.e., x ≠ -8, -6, -4, -2, 4, 6, 8) and see if they satisfy the equation. Since the equation might have integer roots, let's try plugging in some integer values.

Let me list the possible integer roots. Since the equation is a rational function, the roots would be where the numerator equals zero. However, the numerator after combining all terms would be a polynomial, and the roots would be the solutions to that polynomial equation. But without expanding, it's hard to know.

Alternatively, maybe try small integer values like x = 0, 1, 2, 3, 5, 7, etc., and see if they satisfy the equation.

Let's try x = 0:

First term: (0+1)(0-3)/(5*(0+2)(0-4)) = (1)(-3)/(5*2*(-4)) = (-3)/(5*2*(-4)) = (-3)/(-40) = 3/40

Second term: (0+3)(0-5)/(9*(0+4)(0-6)) = (3)(-5)/(9*4*(-6)) = (-15)/(9*4*(-6)) = (-15)/(-216) = 5/72

Third term: 2*(0+5)(0-7)/(13*(0+6)(0-8)) = 2*(5)(-7)/(13*6*(-8)) = 2*(-35)/(13*6*(-8)) = (-70)/( -624 ) = 70/624 = 35/312

Right-hand side: 92/585 ≈ 0.1575

Sum of left-hand side: 3/40 + 5/72 + 35/312

Convert to common denominator, let's see 40, 72, 312. The LCM of 40, 72, 312. 40=8*5, 72=8*9, 312=8*39=8*3*13. So LCM is 8*5*9*13=4680.

Convert each fraction:

3/40 = (3*117)/4680 = 351/4680

5/72 = (5*65)/4680 = 325/4680

35/312 = (35*15)/4680 = 525/4680

Total sum: 351 + 325 + 525 = 1201/4680 ≈ 0.257

Which is not equal to 92/585 ≈ 0.1575. So x=0 is not a root.

How about x=1:

First term: (1+1)(1-3)/(5*(1+2)(1-4)) = (2)(-2)/(5*3*(-3)) = (-4)/( -45 ) = 4/45 ≈ 0.0889

Second term: (1+3)(1-5)/(9*(1+4)(1-6)) = (4)(-4)/(9*5*(-5)) = (-16)/( -225 ) = 16/225 ≈ 0.0711

Third term: 2*(1+5)(1-7)/(13*(1+6)(1-8)) = 2*(6)(-6)/(13*7*(-7)) = 2*(-36)/( -637 ) = (-72)/(-637) ≈ 0.113

Sum: 0.0889 + 0.0711 + 0.113 ≈ 0.273, which is higher than 0.1575. So x=1 is not a root.

x=2:

First term: (2+1)(2-3)/(5*(2+2)(2-4)) = (3)(-1)/(5*4*(-2)) = (-3)/( -40 ) = 3/40 ≈ 0.075

Second term: (2+3)(2-5)/(9*(2+4)(2-6)) = (5)(-3)/(9*6*(-4)) = (-15)/( -216 ) = 15/216 ≈ 0.0694

Third term: 2*(2+5)(2-7)/(13*(2+6)(2-8)) = 2*(7)(-5)/(13*8*(-6)) = 2*(-35)/( -624 ) = (-70)/(-624) ≈ 0.112

Sum: 0.075 + 0.0694 + 0.112 ≈ 0.256, still higher than 0.1575.

x=3:

First term: (3+1)(3-3)/(5*(3+2)(3-4)) = (4)(0)/(5*5*(-1)) = 0/(-25) = 0

Second term: (3+3)(3-5)/(9*(3+4)(3-6)) = (6)(-2)/(9*7*(-3)) = (-12)/( -189 ) = 12/189 ≈ 0.0635

Third term: 2*(3+5)(3-7)/(13*(3+6)(3-8)) = 2*(8)(-4)/(13*9*(-5)) = 2*(-32)/( -585 ) = (-64)/(-585) ≈ 0.1094

Sum: 0 + 0.0635 + 0.1094 ≈ 0.1729, which is still higher than 0.1575.

x=4 is excluded because denominator is zero.

x=5:

First term: (5+1)(5-3)/(5*(5+2)(5-4)) = (6)(2)/(5*7*1) = 12/35 ≈ 0.3429

Second term: (5+3)(5-5)/(9*(5+4)(5-6)) = (8)(0)/(9*9*(-1)) = 0/(-81) = 0

Third term: 2*(5+5)(5-7)/(13*(5+6)(5-8)) = 2*(10)(-2)/(13*11*(-3)) = 2*(-20)/( -429 ) = (-40)/(-429) ≈ 0.0932

Sum: 0.3429 + 0 + 0.0932 ≈ 0.4361, which is way higher.

x=6 is excluded.

x=7:

First term: (7+1)(7-3)/(5*(7+2)(7-4)) = (8)(4)/(5*9*3) = 32/135 ≈ 0.237

Second term: (7+3)(7-5)/(9*(7+4)(7-6)) = (10)(2)/(9*11*1) = 20/99 ≈ 0.202

Third term: 2*(7+5)(7-7)/(13*(7+6)(7-8)) = 2*(12)(0)/(13*13*(-1)) = 0/(-169) = 0

Sum: 0.237 + 0.202 + 0 ≈ 0.439, which is higher.

x=8 is excluded.

How about x= -1:

First term: (-1+1)(-1-3)/(5*(-1+2)(-1-4)) = (0)(-4)/(5*1*(-5)) = 0/( -25 ) = 0

Second term: (-1+3)(-1-5)/(9*(-1+4)(-1-6)) = (2)(-6)/(9*3*(-7)) = (-12)/( -189 ) = 12/189 ≈ 0.0635

Third term: 2*(-1+5)(-1-7)/(13*(-1+6)(-1-8)) = 2*(4)(-8)/(13*5*(-9)) = 2*(-32)/( -585 ) = (-64)/(-585) ≈ 0.1094

Sum: 0 + 0.0635 + 0.1094 ≈ 0.1729, same as x=3.

x= -2 is excluded.

x= -3:

First term: (-3+1)(-3-3)/(5*(-3+2)(-3-4)) = (-2)(-6)/(5*(-1)(-7)) = 12/35 ≈ 0.3429

Second term: (-3+3)(-3-5)/(9*(-3+4)(-3-6)) = (0)(-8)/(9*1*(-9)) = 0/(-81) = 0

Third term: 2*(-3+5)(-3-7)/(13*(-3+6)(-3-8)) = 2*(2)(-10)/(13*3*(-11)) = 2*(-20)/( -429 ) = (-40)/(-429) ≈ 0.0932

Sum: 0.3429 + 0 + 0.0932 ≈ 0.4361

x= -4 is excluded.

x= -5:

First term: (-5+1)(-5-3)/(5*(-5+2)(-5-4)) = (-4)(-8)/(5*(-3)(-9)) = 32/135 ≈ 0.237

Second term: (-5+3)(-5-5)/(9*(-5+4)(-5-6)) = (-2)(-10)/(9*(-1)(-11)) = 20/99 ≈ 0.202

Third term: 2*(-5+5)(-5-7)/(13*(-5+6)(-5-8)) = 2*(0)(-12)/(13*1*(-13)) = 0/(-169) = 0

Sum: 0.237 + 0.202 + 0 ≈ 0.439

x= -6 is excluded.

x= -7:

First term: (-7+1)(-7-3)/(5*(-7+2)(-7-4)) = (-6)(-10)/(5*(-5)(-11)) = 60/275 = 12/55 ≈ 0.218

Second term: (-7+3)(-7-5)/(9*(-7+4)(-7-6)) = (-4)(-12)/(9*(-3)(-13)) = 48/351 ≈ 0.1366

Third term: 2*(-7+5)(-7-7)/(13*(-7+6)(-7-8)) = 2*(-2)(-14)/(13*(-1)(-15)) = 2*28/(195) ≈ 56/195 ≈ 0.2877

Sum: 0.218 + 0.1366 + 0.2877 ≈ 0.6423, which is way higher.

x= -8 is excluded.

Hmm, none of these integer values seem to satisfy the equation. Maybe the roots are not integers. Maybe I need to consider another approach.

Alternatively, perhaps I can set the entire equation equal to zero and combine the terms:

Let me denote:

A = (x+1)(x-3)/[5(x+2)(x-4)]

B = (x+3)(x-5)/[9(x+4)(x-6)]

C = 2(x+5)(x-7)/[13(x+6)(x-8)]

So the equation is A + B - C = 92/585.

Maybe compute A + B - C - 92/585 = 0.

Alternatively, to combine all terms into a single fraction, which is what I was trying earlier. But that's going to be a huge polynomial. Let me see if there's a smarter way.

Alternatively, notice that the denominators are similar. Let me check if multiplying each term by the common denominator will result in a polynomial equation that can be solved.

But since the denominators are all different, the common denominator is the product of all denominators. Let me attempt to compute the numerator.

So, when we multiply both sides by 5*9*13*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8), which is 585*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8), we get:

First term numerator: (x+1)(x-3)*9*13*(x+4)(x-6)(x+6)(x-8)

Second term numerator: (x+3)(x-5)*5*13*(x+2)(x-4)(x+6)(x-8)

Third term numerator: -2(x+5)(x-7)*5*9*(x+2)(x-4)(x+6)(x-8)

Set equal to 92/585 * common denominator, which is 92*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8)

So bringing everything to one side:

First term + second term - third term - 92*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8) = 0

Let me denote this as:

Numerator = [ (x+1)(x-3)*9*13*(x+4)(x-6)(x+6)(x-8) + (x+3)(x-5)*5*13*(x+2)(x-4)(x+6)(x-8) - 2(x+5)(x-7)*5*9*(x+2)(x-4)(x+6)(x-8) - 92*(x+2)(x-4)(x+4)(x-6)(x+6)(x-8) ] = 0

This is a polynomial equation of degree 6, which seems very difficult to solve directly. However, maybe there's a way to factor this polynomial or find roots through substitution.

Alternatively, perhaps notice that the equation can be transformed by substituting y = x + something. Let me check the symmetry.

Looking at the original equation, the denominators and numerators have a sort of symmetry with shifts of 2. For example, the denominators are (x+2)(x-4), (x+4)(x-6), (x+6)(x-8). Each denominator is (x + 2k)(x - (2k + 2)) for k=1,2,3. Similarly, the numerators are (x + (2k -1))(x - (2k +1)) for k=1,2,3.

This suggests that there might be a substitution that can simplify the equation. Let me consider substituting z = x + 4. Then x = z - 4. Let's see how the terms transform.

First term:

(x+1)(x-3) = (z - 4 +1)(z -4 -3) = (z -3)(z -7)

Denominator: 5(x+2)(x-4) = 5(z -4 +2)(z -4 -4) = 5(z -2)(z -8)

Second term:

(x+3)(x-5) = (z -4 +3)(z -4 -5) = (z -1)(z -9)

Denominator: 9(x+4)(x-6) = 9(z -4 +4)(z -4 -6) = 9(z)(z -10)

Third term:

2(x+5)(x-7) = 2(z -4 +5)(z -4 -7) = 2(z +1)(z -11)

Denominator:13(x+6)(x-8) =13(z -4 +6)(z -4 -8) =13(z +2)(z -12)

Right-hand side: 92/585

So substituting z = x +4, the equation becomes:

[ (z -3)(z -7) / [5(z -2)(z -8) ] ] + [ (z -1)(z -9) / [9z(z -10) ] ] - [ 2(z +1)(z -11) / [13(z +2)(z -12) ] ] = 92/585

Hmm, not sure if this helps. Let me check the denominators:

First term denominator: 5(z -2)(z -8)

Second term denominator: 9z(z -10)

Third term denominator:13(z +2)(z -12)

Still complex. Maybe another substitution? Alternatively, perhaps shifting z again. Let me try substituting w = z - something. Maybe shift to center around zero.

Alternatively, perhaps note that the terms are similar but shifted. Let me see the expressions:

First term: (z -3)(z -7) / [5(z -2)(z -8) ]

Second term: (z -1)(z -9) / [9z(z -10) ]

Third term: 2(z +1)(z -11) / [13(z +2)(z -12) ]

Notice that the first term's numerator is (z -3)(z -7), and denominator is (z -2)(z -8). Similarly, the second term's numerator is (z -1)(z -9), denominator (z)(z -10). Third term is 2*(z +1)(z -11) over (z +2)(z -12).

Wait, if we look at the first term, numerator is (z -3)(z -7), denominator is (z -2)(z -8). If we factor these, it's close to (z - (3))(z - (7)) over (z - (2))(z - (8)). Similarly, the second term is (z -1)(z -9) over z(z -10). Third term is (z +1)(z -11) over (z +2)(z -12).

Hmm, each numerator is (z - a)(z - (a +4)) and denominator is (z - b)(z - (b +6)), where a and b are constants. Maybe there's a pattern here.

Alternatively, perhaps the substitution t = z - 5? Let's try that.

Let t = z -5, so z = t +5. Then:

First term numerator: (t +5 -3)(t +5 -7) = (t +2)(t -2)

First term denominator: (t +5 -2)(t +5 -8) = (t +3)(t -3)

Second term numerator: (t +5 -1)(t +5 -9) = (t +4)(t -4)

Second term denominator: (t +5)(t +5 -10) = (t +5)(t -5)

Third term numerator: 2(t +5 +1)(t +5 -11) = 2(t +6)(t -6)

Third term denominator:13(t +5 +2)(t +5 -12) =13(t +7)(t -7)

So substituting t = z -5, the equation becomes:

[ (t +2)(t -2) / [5(t +3)(t -3) ] ] + [ (t +4)(t -4) / [9(t +5)(t -5) ] ] - [ 2(t +6)(t -6) / [13(t +7)(t -7) ] ] = 92/585

Hmm, this looks symmetric. Let's note that the first term is (t^2 -4)/[5(t^2 -9)], the second term is (t^2 -16)/[9(t^2 -25)], the third term is 2(t^2 -36)/[13(t^2 -49)], and the right-hand side is 92/585.

So the equation is:

[ (t^2 -4)/(5(t^2 -9)) ] + [ (t^2 -16)/(9(t^2 -25)) ] - [ 2(t^2 -36)/(13(t^2 -49)) ] = 92/585

This seems more manageable. Let me denote u = t^2. Then, the equation becomes:

[ (u -4)/(5(u -9)) ] + [ (u -16)/(9(u -25)) ] - [ 2(u -36)/(13(u -49)) ] = 92/585

This is a rational equation in u. Let me write it as:

A + B - C = D, where:

A = (u -4)/(5(u -9))

B = (u -16)/(9(u -25))

C = 2(u -36)/(13(u -49))

D = 92/585

Let me combine terms:

A + B - C - D = 0

So:

(u -4)/(5(u -9)) + (u -16)/(9(u -25)) - 2(u -36)/(13(u -49)) - 92/585 = 0

To combine these terms, we need a common denominator. The denominators are 5(u -9), 9(u -25), 13(u -49), and 585. Note that 585 = 5*9*13, so the common denominator is 585(u -9)(u -25)(u -49).

Multiply each term by the common denominator:

First term: (u -4)*9*13*(u -25)(u -49)

Second term: (u -16)*5*13*(u -9)(u -49)

Third term: -2(u -36)*5*9*(u -9)(u -25)

Fourth term: -92/585 * 585(u -9)(u -25)(u -49) = -92(u -9)(u -25)(u -49)

So the equation becomes:

First term + Second term + Third term + Fourth term = 0

Let me compute each term:

First term: 9*13*(u -4)*(u -25)*(u -49) = 117*(u -4)(u^2 -74u + 1225)

Second term: 5*13*(u -16)*(u -9)*(u -49) = 65*(u -16)(u^2 -58u + 441)

Third term: -2*5*9*(u -36)*(u -9)*(u -25) = -90*(u -36)(u^2 -34u + 225)

Fourth term: -92*(u -9)(u -25)(u -49)

So expanding each term:

First term:

117*(u -4)(u^2 -74u +1225)

= 117*[u*(u^2 -74u +1225) -4*(u^2 -74u +1225)]

= 117*[u^3 -74u^2 +1225u -4u^2 +296u -4900]

= 117*u^3 -117*78u^2 +117*(1225 +296)u -117*4900

Wait, let me compute each part step by step.

First, expand (u -4)(u^2 -74u +1225):

= u*(u^2 -74u +1225) -4*(u^2 -74u +1225)

= u^3 -74u^2 +1225u -4u^2 +296u -4900

= u^3 -78u^2 +1521u -4900

Multiply by 117:

117u^3 -78*117u^2 +1521*117u -4900*117

Compute coefficients:

78*117: 70*117=8190, 8*117=936, total 8190+936=9126

1521*117: Let's compute 1500*117=175,500, 21*117=2,457, total 175,500 +2,457=177,957

4900*117=4900*100 +4900*17=490,000 +83,300=573,300

So first term expanded:

117u^3 -9126u^2 +177,957u -573,300

Second term:

65*(u -16)(u^2 -58u +441)

First, expand (u -16)(u^2 -58u +441):

= u*(u^2 -58u +441) -16*(u^2 -58u +441)

= u^3 -58u^2 +441u -16u^2 +928u -7056

= u^3 -74u^2 +1369u -7056

Multiply by 65:

65u^3 -74*65u^2 +1369*65u -7056*65

Calculate coefficients:

74*65=4,810

1369*65: 1300*65=84,500, 69*65=4,485, total 84,500 +4,485=88,985

7056*65: 7000*65=455,000, 56*65=3,640, total 455,000 +3,640=458,640

So second term expanded:

65u^3 -4,810u^2 +88,985u -458,640

Third term:

-90*(u -36)(u^2 -34u +225)

First, expand (u -36)(u^2 -34u +225):

= u*(u^2 -34u +225) -36*(u^2 -34u +225)

= u^3 -34u^2 +225u -36u^2 +1,224u -8,100

= u^3 -70u^2 +1,449u -8,100

Multiply by -90:

-90u^3 +70*90u^2 -1,449*90u +8,100*90

Calculate coefficients:

70*90=6,300

1,449*90=130,410

8,100*90=729,000

So third term expanded:

-90u^3 +6,300u^2 -130,410u +729,000

Fourth term:

-92*(u -9)(u -25)(u -49)

First, expand (u -9)(u -25)(u -49). Let's do this step by step.

First, multiply (u -9)(u -25):

= u^2 -25u -9u +225 = u^2 -34u +225

Then multiply by (u -49):

= (u^2 -34u +225)(u -49)

= u^3 -49u^2 -34u^2 +1,666u +225u -11,025

= u^3 -83u^2 +1,891u -11,025

Multiply by -92:

-92u^3 +83*92u^2 -1,891*92u +11,025*92

Calculate coefficients:

83*92=7,636

1,891*92: 1,800*92=165,600, 91*92=8,372, total 165,600 +8,372=173,972

11,025*92: 10,000*92=920,000, 1,025*92=94,300, total 920,000 +94,300=1,014,300

So fourth term expanded:

-92u^3 +7,636u^2 -173,972u +1,014,300

Now, combining all four terms:

First term: 117u^3 -9126u^2 +177,957u -573,300

Second term: 65u^3 -4,810u^2 +88,985u -458,640

Third term: -90u^3 +6,300u^2 -130,410u +729,000

Fourth term: -92u^3 +7,636u^2 -173,972u +1,014,300

Combine like terms:

u^3: 117 +65 -90 -92 = 117+65=182; 182 -90=92; 92 -92=0

u^2: -9126 -4,810 +6,300 +7,636

= (-9126 -4,810) + (6,300 +7,636)

= (-13,936) +13,936 = 0

u terms: 177,957 +88,985 -130,410 -173,972

= (177,957 +88,985) - (130,410 +173,972)

= 266,942 -304,382 = -37,440

Constant terms: -573,300 -458,640 +729,000 +1,014,300

= (-573,300 -458,640) + (729,000 +1,014,300)

= (-1,031,940) +1,743,300 = 711,360

So overall, combining all terms:

0u^3 +0u^2 -37,440u +711,360 = 0

Simplify:

-37,440u +711,360 = 0

Solve for u:

-37,440u = -711,360

u = (-711,360)/(-37,440) = 711,360 /37,440

Divide numerator and denominator by 480: 711,360 ÷480=1,482; 37,440 ÷480=78

So 1,482 /78 = 19

Therefore, u =19.

Recall that u = t^2, and t = z -5.

So:

t^2 =19 => t = sqrt(19) or t = -sqrt(19)

t = z -5, so z = t +5

Therefore:

z =5 + sqrt(19) or z=5 - sqrt(19)

But z =x +4, so x = z -4

Thus:

x =5 + sqrt(19) -4 =1 + sqrt(19)

or

x=5 - sqrt(19) -4 =1 - sqrt(19)

Therefore, the real roots are x=1 + sqrt(19) and x=1 - sqrt(19)

Let me verify if these are valid solutions, i.e., they are not excluded values (which were x=-8,-6,-4,-2,4,6,8). Since sqrt(19) is approximately 4.358, so 1 + sqrt(19) ≈5.358, and 1 - sqrt(19)≈-3.358. Neither of these is excluded, so they are valid solutions.

Therefore, the real roots are x=1 + sqrt(19) and x=1 - sqrt(19).

**Final Answer**
The real roots are \boxed{1 + \sqrt{19}} and \boxed{1 - \sqrt{19}}.
Therefore, the real roots of the given equation are $\boxed{1 + \sqrt{19}}$ and $\boxed{1 - \sqrt{19}}$.
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