Quadratic powered

Case 1: For some non-zero real number $a$ , $(x^2-2x)^{x^2+x-6}=a^0$ $\implies x^2+x-6=(x+3)(x-2)=0$ When $x=2$ , our expression attains the form $0^0$ which is indeterminate. So for this case, the only solution is $\boxed{x=-3}$

Case 2: For some real number $a$ , $(x^2-2x)^{x^2+x-6}=1^a$ $\implies x^2-2x=1$ $\implies (x-(1+\sqrt{2}))(x-(1-\sqrt{2}))=0$ This leads us to the solutions $\boxed{x=1+\sqrt{2}}$ and $\boxed{x=1-\sqrt{2}}$

Case 3: For some even $a$ , $(x^2-2x)^{x^2+x-6}=(-1)^{a}$ $\implies x^2-2x=-1$ $\implies (x-1)^2=0$ This gives us our final solution $\boxed{x=1}$

Therefore, the sum of all real values $x$ that satisfy the given equation is $(-3)+(1+\sqrt{2})+(1-\sqrt{2})+(1)=\boxed{0}$