Answer
$\{3, \frac{21}{5}, \frac{24}{5}, 7\}$
Work Step by Step
Step 1. Let $u=(x-5)^{2}$, we have $25u^2-116u+64=0$
Step 2. Factor and solve: $(u-4)(25u-16)=0$ thus $u=4,\frac{16}{25}$
Step 3. For $u=4$, we have $(x-5)^{2}=4$ or $x-5=\pm2$ thus $x=3,7$
Step 4. For $u=\frac{16}{25}$, we have $(x-5)^{2}=\frac{16}{25}$ or $x-5=\pm\frac{4}{5}$ thus $x=\frac{21}{5}, \frac{24}{5}$
Step 5. Solution set $\{3, \frac{21}{5}, \frac{24}{5}, 7\}$
