Answer
$\{-3,0\}$
Work Step by Step
Use the square root method by taking the square root of both sides.
$\sqrt{(2y+3)^2}=\pm\sqrt {9}$
$2y+3=\pm\sqrt {3^2}$
$2y+3=\pm 3$
Split the expression to obtain:
$2y+3 =3\quad $ or $\quad 2y+3=-3$
Subtract $3$ from both sides of both equations.
$2y+3 -3=3-3\quad $ or $\quad 2y+3-3=-3-3$
Simplify.
$2y=0\quad$ or $\quad 2y=-6$
Divide both sides of both equations by $2$.
$y=0\quad $ or $\quad y=-3$
Hence, the solution set is $\{-3,0\}$.
