Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 592: 15

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

Given: $2(x+2)^2=16$ This can be written as: $(x+2)=8$ Apply The Square Root Property. or, $x+2=\sqrt{8}$ or, $x+2=\pm (2\sqrt 2)$ or, $x=-2+2\sqrt 2,-2-2\sqrt 2$ Hence, solution set is $x=${$-2+2\sqrt 2,-2-2\sqrt 2$}