jmc

We start by completing the square: $$\begin{array}{rl}-x^2-8x+12& =-(x^2+8x)+12\\ & =-(x^2+8x+(8/2{)}^2-(8/2{)}^2)+12\\ & =-((x+4{)}^2-4^2)+12\\ & =-(x+4{)}^2+4^2+12\\ & =-(x+4{)}^2+28.\end{array}$$ 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 $x=-4$.