## Algebra 2 (1st Edition)

The solutions are $2+\sqrt{\frac{40}{3}}$ and $2-\sqrt{\frac{40}{3}}$.
$3(x-2)^{2}=40\qquad$ ...divide each side with $3$. $(x-2)^{2}=\displaystyle \frac{40}{3}\qquad$ ...take square roots of each side. $\sqrt{(x-2)^{2}}=\sqrt{\frac{40}{3}}\qquad$ ...simplify. $x-2=\pm\sqrt{\frac{40}{3}}\qquad$ ...add $2$ to each side. $x-2+2=\pm\sqrt{\frac{40}{3}}+2\qquad$ ...simplify. $x=2\pm\sqrt{\frac{40}{3}}$