jmc

We start by completing the square: $$\begin{array}{rl}-x^2-6x+12& =-(x^2+6x)+12\\ & =-(x^2+6x+(6/2{)}^2-(6/2{)}^2)+12\\ & =-((x+3{)}^2-3^2)+12\\ & =-(x+3{)}^2+3^2+12\\ & =-(x+3{)}^2+21.\end{array}$$Since the square of a real number is at least 0, we have $(x+3{)}^2\ge 0$, so $-(x+3{)}^2\le 0$. Therefore, $-(x+3{)}^2+21$ is at most 21. Since $(x+3{)}^2=0$ when $x=-3$, this maximum of $21$ is achieved when $x=\boxed{-3}$.