Answer
$x=-2$ or $x=1$
Work Step by Step
We solve:
$(2x+1)^{2}=9$
We take the square root of both sides:
$\sqrt{(2x+1)^{2}}=\pm\sqrt{9}$
$2x+1=\pm\sqrt{9}$
$2x+1=\pm 3$
$2x=-1\pm 3$
$x=\frac{-1\pm 3}{2}$
$x=-\frac{4}{2}=-2$ or $x=\frac{2}{2}=1$
$x=-2$ or $x=1$