## Intermediate Algebra (6th Edition)

The solution set is $\left\{2-3\sqrt3, 2+3\sqrt3\right\}$.
Get the square root of both sides: $\sqrt{(x-2)^2} \pm \sqrt{27} \\x-2 = \pm \sqrt{9(3)} \\x-2 = \pm 3\sqrt{3}$ Add 2 to both sides: $x-2+2 = 2 \pm 3\sqrt{3} \\x=2 \pm 3\sqrt{3}$