jmc

We complete the square: $$\begin{array}{rl}-5r^2+40r-12& =(-5r^2+40r)-12\\ & =-5(r^2-8r+16)-12+5\cdot 16\\ & =-5(r-4{)}^2+68\end{array}$$ The maximum value of $-5(r-4{)}^2$ is $0$, since the square of a real number is never negative. Thus, the maximum value of the expression is $\boxed{68}$.