## Precalculus (10th Edition)

$x=5$ or $x=-1$
Take the square root of both sides \begin{align*} \sqrt{(x-2)^2}&=\pm \sqrt{9}\\ x-2&=\pm 3\\ x&=2 \pm 3 \end{align*} Hence, $x=2+3=5$ or $x=2-3=-1$