Answer
$\{0,\frac{4}{3}\}$
Work Step by Step
Use the square root method by taking the square root of both sides.
$\sqrt{(3x-2)^2}=\pm\sqrt {4}$
$3x-2=\pm\sqrt {2^2}$
$3x-2=\pm2$
Split the expression to obtain:
$3x-2 =2\quad $ or $\quad 2y+3=-2$
Add $2$ to both sides of both equations.
$3x-2 +2=2+2$ or $3x-2+2=-2+2$
Simplify.
$3x=4\quad $ or $\quad 3x=0$
Divide both sides by $3$.
$x=\frac{4}{3}$ or $x=0$
Hence, the solution set is $\{0,\frac{4}{3}\}$.
