Answer
$\{-2\pm\frac{\sqrt 2}{2}, -2\pm\frac{2\sqrt 3}{3} \}$
Work Step by Step
Step 1. Let $u=(x+2)^{2}$, we have $6u^2-11u+4=0$
Step 2. Factor and solve: $(2u-1)(3u-4)=0$ thus $u=\frac{1}{2}, \frac{4}{3}$
Step 3. For $u=\frac{1}{2}$, we have $(x+2)^{2}=\frac{1}{2}$ or $x+2=\pm\frac{\sqrt 2}{2}$ thus $x=-2\pm\frac{\sqrt 2}{2}$
Step 4. For $u=\frac{4}{3}$, we have $(x+2)^{2}=\frac{4}{3}$ or $x+2=\pm\frac{2\sqrt 3}{3}$ thus $x=-2\pm\frac{2\sqrt 3}{3}$
Step 5. Solution set $\{-2\pm\frac{\sqrt 2}{2}, -2\pm\frac{2\sqrt 3}{3} \}$
