#### Answer

$\color{blue}{\left\{-2, 1\right\}}$.

#### Work Step by Step

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 $\color{blue}{\left\{-2, 1\right\}}$.