Answer
$\frac{9}{2},\frac{11}{2}, \frac{14}{3},\frac{16}{3}$
Work Step by Step
Step 1. Let $u=(x-5)^{-2}$, the original equation becomes $u^2-13u=-36$ or $u^2-13u+36=0$
Step 2. Factor the equation to get $(u-4)(u-9)=0$, thus $x=4$ and $x=9$
Step 3. For $u=(x-5)^{-2}=4$, we have $(x-5)^2=\frac{1}{4}\longrightarrow x=5\pm\frac{1}{2}$, thus $x=\frac{9}{2}$ and $x=\frac{11}{2}$
Step 4. For $u=(x-5)^{-2}=9$, we have $(x-5)^2=\frac{1}{9}\longrightarrow x=5\pm\frac{1}{3}$, thus $x=\frac{14}{3}$ and $x=\frac{16}{3}$
Step 5. The solutions are $x=\frac{9}{2},\frac{11}{2}, \frac{14}{3},\frac{16}{3}$
