## College Algebra (10th Edition)

$x=\left\{-1, 5\right\}$
Take the square root of both sides to obtain: $\sqrt{(x-2)^2}=\pm\sqrt9 \\x-2 = \pm 3$ Add $2$ on both sides to obtain: $x=2\pm3$ Thus, $x_1 =2+3=5 \\x_1=2-3=-1$ Therefore, the solutions are: $x=\left\{-1, 5\right\}$