<Question>
For what values of $x$ is it true that $x^2 - 5x - 4 \le 10$? Express your answer in interval notation.
</Question>

<think>
<Plan_1>
Okay, so I need to find all the values of x where x squared minus 5x minus 4 is less than or equal to 10. Hmm, let me start by writing down the inequality: x² - 5x - 4 ≤ 10. 

First, maybe I should move all terms to one side to make it easier to handle. If I subtract 10 from both sides, that would give me x² - 5x - 4 - 10 ≤ 0.
Simplifying that, it becomes x² - 5x - 14 ≤ 0.

Alright, so now I have a quadratic inequality: x² - 5x - 14 ≤ 0. To solve this, I remember that I need to find the roots of the quadratic equation first, which are the points where the quadratic equals zero. Then, I can determine the intervals where the quadratic is negative or zero.

Let me try to factor the quadratic. The equation is x² - 5x - 14 = 0. To factor this, I need two numbers that multiply to -14 and add up to -5. Let's see... The factors of 14 are 1 and 14, 2 and 7. Since the product is negative, one number has to be positive and the other negative.

<prompt_1.1> Factor the quadratic equation x² - 5x - 14 = 0 and find its roots. </prompt_1.1>
-----

Hmm, but let me double-check my steps to make sure I didn't make a mistake.

Wait, let me verify the factoring again.

<prompt_1.2> Verify the factoring of the quadratic equation x² - 5x - 14 = 0. </prompt_1.2>
-----

But just to be thorough, let me test another point in each interval again.

<prompt_1.3> Test another point in each interval to verify the solution of the inequality. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's try 2 and -7: 2 * (-7) = -14, and 2 + (-7) = -5. Yes, that works. So the quadratic factors as (x + 2)(x - 7) = 0. Therefore, the roots are x = -2 and x = 7.
</execution_1.1>

<execution_1.2>
If I take -7 and +2, their sum is -5 and product is -14. Yes, that's correct. So the roots are x = -2 and x = 7. Wait, hold on. If the factors are (x - 7)(x + 2) = 0, then the solutions are x = 7 and x = -2. That's correct.
</execution_1.2>

<execution_1.3>
For x < -2, say x = -3: (-3)^2 -5*(-3) -14 = 9 +15 -14 = 10, which is not ≤0. So that's correct. 

Between -2 and 7, say x = 0: 0 -0 -14 = -14 ≤0. Correct. 

For x >7, say x=8: 64 -40 -14=10, which is again not ≤0.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, the quadratic can be written as (x + 2)(x - 7) ≤ 0. 

Next, I need to determine where this product is less than or equal to zero. The critical points are x = -2 and x = 7, which divide the real number line into three intervals: (-∞, -2), (-2, 7), and (7, ∞). 

To test each interval, I can pick a test point in each interval and see if the product is positive or negative there.

<prompt_2.1> Determine the intervals where the quadratic expression (x + 2)(x - 7) is less than or equal to zero. </prompt_2.1>
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But wait, let me confirm by plugging in numbers. Let's take x = -2: (-2)^2 -5*(-2) -14 = 4 +10 -14 = 0.

<prompt_2.2> Confirm the solution by testing the boundary points x = -2 and x = 7. </prompt_2.2>
-----

But let me also think about the graph of the quadratic. Since the coefficient of x² is positive (1 in this case), the parabola opens upwards. Therefore, the graph will be below or equal to the x-axis (i.e., y ≤ 0) between the two roots.

<prompt_2.3> Analyze the graph of the quadratic x² - 5x - 14 to confirm the solution. </prompt_2.3>

I think that's solid. Let me just recap the steps to ensure I didn't skip anything:

<prompt_2.4> Recap the steps taken to solve the inequality and confirm the solution. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
First interval: (-∞, -2). Let's take x = -3. Plugging into (x + 2)(x - 7): (-3 + 2)(-3 - 7) = (-1)(-10) = 10, which is positive. So this interval is not part of the solution. 

Second interval: (-2, 7). Let's take x = 0. Plugging in: (0 + 2)(0 - 7) = (2)(-7) = -14, which is negative. So this interval is part of the solution. 

Third interval: (7, ∞). Let's take x = 8. Plugging in: (8 + 2)(8 - 7) = (10)(1) = 10, which is positive. So not part of the solution.
</execution_2.1>

<execution_2.2>
So, x = -2 is included. For x = 7: 49 - 35 -14 = 0. So x =7 is also included. For x = 0 (between -2 and 7): 0 -0 -14 = -14, which is ≤0. For x = -3 (left of -2): 9 +15 -14 = 10, which is not ≤0. For x = 8 (right of 7): 64 -40 -14 = 10, also not ≤0.
</execution_2.2>

<execution_2.3>
That makes sense why the interval is between -2 and 7. So that confirms again the solution.

Alternatively, if I recall, for a quadratic ax² + bx + c with a > 0, the quadratic is ≤ 0 between its roots. If there are real roots. If the quadratic doesn't cross the x-axis (discriminant negative), then it's always positive or always negative. But here, since there are two real roots, it's negative between them.

So this is another way to confirm. The quadratic x² -5x -14 opens upwards, so it's below zero between its roots. Hence, the solution is [-2,7].
</execution_2.3>

<execution_2.4>
1. Started with the inequality x² -5x -4 ≤ 10.
2. Subtracted 10 from both sides to get x² -5x -14 ≤ 0.
3. Factored the quadratic into (x + 2)(x -7) = 0, giving roots at x = -2 and x =7.
4. Test intervals around the roots to determine where the quadratic is negative or zero.
5. Concluded that the solution is the interval between the roots, including the endpoints since the inequality is non-strict (≤).
6. Verified with test points and considered the graph of the quadratic to confirm the solution.

Everything seems to check out. I don't see any mistakes in the reasoning.
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So the solution set is the interval [-2, 7]. But wait, let me make sure I didn't make a mistake in the sign when testing the intervals. 

For the first interval, when x is less than -2, the product (x + 2)(x - 7) is positive because both factors are negative: (negative)(negative) = positive. 

For the middle interval, between -2 and 7, (x + 2) is positive and (x - 7) is negative, so their product is negative. 

For the last interval, x greater than 7, both factors are positive: (positive)(positive) = positive. 

Since the inequality is looking for where the product is less than or equal to zero, that's only the middle interval where the product is negative. And since the inequality includes equality (≤ 0), we include the endpoints where the expression equals zero.
-----

Based on execution_2.2:
And x =0 is in the interval, and it's negative there. So that's correct.
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the solution in interval notation is $\boxed{[-2,7]}$.
</Final_answer>
</think>