## Precalculus (6th Edition)

$\color{blue}{\left\{-2, 1\right\}}$.
Take the square root of both sides to obtain: $\sqrt{(2x+1)^2}=\pm \sqrt{9} \\2x+1=\pm 3$ Subtract 1 to both sides: $2x = \pm 3 - 1 \\2x = -1 \pm 3$ Divide $2$ to both sides: $\dfrac{2x}{2} = \dfrac{-1\pm 3}{2} \\x = -\dfrac{1}{2} \pm \dfrac{3}{2} \\x_1=-\frac{1}{2}-\frac{3}{2}=-\frac{4}{2}=-2 \\x_2=-\frac{1}{2}+\frac{3}{2}=\frac{2}{2} = 1$ Thus, the solution set is $\color{blue}{\left\{-2, 1\right\}}$.