problem: What is the maximum value of $-x^2- 8x + 12$?
solution: We start by completing the square: \begin{align*}
  -x^2 -8x +12 &= -(x^2 + 8x) + 12\\
  & = -(x^2 + 8x + (8/2)^2 - (8/2)^2) + 12\\
  & = -((x+4)^2 -4^2) + 12 \\
  &= -(x+4)^2 +4^2 + 12 \\
  &= -(x+4)^2 + 28.\end{align*} Since the square of a real number is at least 0, we have $(x+4)^2\ge 0$, so $-(x+4)^2 \le 0$.  Therefore, $-(x+4)^2 + 28$ is at most 28.  Since $(x+4)^2 =0$ when $x=-4$, this maximum of $\boxed{28}$ is achieved when $x= -4$.