Question:
The quadratic $3x^2-24x+72$ can be written in the form $a(x+b)^2+c$, where $a$, $b$, and $c$ are constants. What is $a+b+c$?

Answer:
We complete the square.

Factoring $3$ out of the quadratic and linear terms gives $3x^2 - 24x = 3(x^2 - 8x)$.

Since $(x-4)^2 = x^2 - 8x + 16$, we can write $$3(x-4)^2 = 3x^2 - 24x + 48.$$This quadratic agrees with the given $3x^2-24x+72$ in all but the constant term. We can write

\begin{align*}
3x^2 - 24x + 72 &= (3x^2 - 24x + 48) + 24 \\
&= 3(x-4)^2 + 24.
\end{align*}Therefore, $a=3$, $b=-4$, $c=24$, and $a+b+c = 3-4+24 = \boxed{23}$.