College Algebra (10th Edition)

The solution set is {$-7,-1$}
To find the real solutions of $(x+4)^2=9$ with the Square Root Method, we take the square root of both sides and then solve for x. $\sqrt{(x+4)^2}=\sqrt{9}$ $x+4=\pm3$ $x=-4\pm3$ Therefore, the solution set is {$-7,-1$}