Answer
$\{4\pm\frac{\sqrt 2}{2}, 4\pm\frac{\sqrt 3}{2} \}$
Work Step by Step
Step 1. Let $u=(x-4)^{2}$, we have $8u^2-10u+3=0$
Step 2. Factor and solve: $(2u-1)(4u-3)=0$ thus $u=\frac{1}{2}, \frac{3}{4}$
Step 3. For $u=\frac{1}{2}$, we have $(x-4)^{2}=\frac{1}{2}$ or $x-4=\pm\frac{\sqrt 2}{2}$ thus $x=4\pm\frac{\sqrt 2}{2}$
Step 4. For $u=\frac{3}{4}$, we have $(x-4)^{2}=\frac{3}{4}$ or $x-4=\pm\frac{\sqrt 3}{2}$ thus $x=4\pm\frac{\sqrt 3}{2}$
Step 5. Solution set $\{4\pm\frac{\sqrt 2}{2}, 4\pm\frac{\sqrt 3}{2} \}$