Let $u=x^2-8x$. We have $u+20=2\sqrt{u+30} \rightarrow u^2+40u+400=4u+120 \rightarrow u^2-36u+280=0 \rightarrow u = 18 \pm 2\sqrt{11}$. Remember that this is equal to $x^2-8x$. With $x^2-8x+\left(18+2\sqrt{11}\right)$ it is easy to see that the discriminant is negative, so no real solutions. With $x^2-8x+\left(18-2\sqrt{11}\right)=0$, the discriminant is positive (real solutions!) so the sum of the roots is $\boxed{8}$.