jmc

Writing the right hand side with 2 as the base, we have $4^{x-4}=(2^2{)}^{x-4}=2^{2(x-4)}=2^{2x-8}$, so our equation is $$2^{x^2-3x-2}=2^{2x-8}.$$Then, by setting the exponents equal to each other, we obtain $$x^2-3x-2=2x-8.$$This gives the quadratic $$x^2-5x+6=0.$$Factoring gives $(x-2)(x-3)=0$, which has solutions $x=2,3$. The sum of these solutions is $\boxed{5}$.