## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\left\{-1,5\right\}$
Apply square root property by taking the square root of both sides: $\sqrt{(x-2)^2}=\pm \sqrt{9}\\ x-2= \pm 3\\ x=2\pm 3$ Thus, $x_1=2+3=5\\ x_2=2-3=-1$