Answer
$\{-4,-3,0,1\}$
Work Step by Step
Recall that $|a|=b \implies a=b$ or $a=-b$.
Thus, using the rule above yields:
$x^2+3x-2=2\quad $ or $\quad x^2+3x-2=-2$
Solve each equation.
Solve $x^2+3x-2=2$:
Subtract $2$ from both sides.
$x^2+3x-2-2=2-2$
Simplify.
$x^2+3x-4=0$
Rewrite $3x$ as $4x-x$.
$x^2+4x-x-4=0$
Group the terms.
$(x^2+4x)+(-x-4)=0$
Factor each group.
$x(x+4)-1(x+4)=0$
Factor out $(x+4)$.
$(x+4)(x-1)=0$
Use Zero-Product Property.
$x+4=0\quad$ or $\quad x-1=0$
Solve for $x$.
$x=-4\quad $ or $\quad x=1$
Solve $x^2+3x-2=-2$:
Add $2$ to both sides.
$x^2+3x-2+2=-2+2$
Simplify.
$x^2+3x=0$
Factor out $x$.
$x(x+3)=0$
Use Zero-Product Property.
$x=0$ or $x+3=0$
Solve for $x$.
$x=0$ or $x=-3$
Hence, the solution set of the given equation is $\{-4,-3,0,1\}$.
