Answer
$\left\{\dfrac{-3-2\sqrt{2}}{2},\dfrac{-3+2\sqrt{2}}{2}\right\}$
Work Step by Step
Taking the square root of both sides (Square Root Property), the given equation, $
(2x+3)^2=8
,$ is equivalent to
\begin{align*}
2x+3=\pm\sqrt{8}
.\end{align*}
Using the properties of radicals and of equality, the equation above is equivalent to
\begin{align*}
2x+3&=\pm\sqrt{4\cdot2}
\\
2x+3&=\pm\sqrt{4}\cdot\sqrt{2}
\\
2x+3&=\pm2\sqrt{2}
\\
2x&=-3\pm2\sqrt{2}
\\\\
x&=\dfrac{-3\pm2\sqrt{2}}{2}
.\end{align*}
Hence, the solution set of the equation $
(2x+3)^2=8
$ is $\left\{\dfrac{-3-2\sqrt{2}}{2},\dfrac{-3+2\sqrt{2}}{2}\right\}$.
