jmc

The area of the square is $(x-2{)}^2$, while the area of the rectangle is $(x-3)(x+4)$. Now we set the area of the rectangle equal to twice the area of the square and solve for $x$: $$\begin{array}{rl}2(x-2{)}^2& =(x-3)(x+4)\Rightarrow \\ 2(x^2-4x+4)& =(x^2+x-12)\Rightarrow \\ x^2-9x+20& =0\Rightarrow \\ (x-5)(x-4)& =0.\end{array}$$So we know that $x=5$ or $x=4$, and our answer is $5+4=\boxed{9}$.