Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 592: 16

{$-2+2\sqrt 3,-2-2\sqrt 3$}

Given: $3(x+2)^2=36$ This can be written as: $(x+2)=12$ Apply The Square Root Property. or, $x+2=\sqrt{12}$ or, $x+2=\pm (2\sqrt 3)$ or, $x=-2+2\sqrt 3,-2-2\sqrt 3$ Hence, solution set is $x=${$-2+2\sqrt 3,-2-2\sqrt 3$}