Answer
The solution is $\left\{-3, 0\right\}$.
Work Step by Step
Take the square root of both sides to obtain:
$2y+3 = \pm \sqrt{9}
\\2y+3 = \pm \sqrt{3^2}
\\2y+3 = \pm 3$
Subtract 3 to both sides to obtain:
$2y = \pm 3 - 3
\\2y = -3 \pm 3$
Divide 2 on both sides of the equation to obtain:
$y=\dfrac{-3 \pm 3}{2}$
Split the solutions to obtain:
$y_1 = \dfrac{-3+3}{2} =0
\\y_2 = \dfrac{-3-3}{2} = -3$
Thus, the solution is $\left\{-3, 0\right\}$.
