Answer
The solution set is $\left\{-1, -\frac{2}{3}\right\}$.
Work Step by Step
RECALL:
A quadratic trinomial $ax^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = ac$ and $d+e =b$
If such integers exist, then $ax^2+bx + c = ax^2+dx + ex + c$, and may be factored by grouping.
The trinomial in the equation has $a=3$, $b=5$, and $c=2$.
Thus, $ac = 3(2) = 6$
The factors of $6$ whose sum is equal to the coefficient of the middle term (which is $b=5$) are $3$ and $2$.
This means that $d=3$ and $e=2$.
Rewrite the middle term of the trinomial as $3x$ +$2x$ to obtain:
$3x^2+5x+2 = 0
\\3x^2+3x+2x+2 = 0$
Regroup then factor out the GCF in each group to obtain:
$(3x^2+3x)+(2x+2)=0
\\3x(x+1) +2(x+1)=0$
Factor out $x+1$ to obtain:
$(x+1)(3x+2)=0$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
$\begin{array}{ccc}
&x+1=0 &\text{ or } &3x+2=0
\\&x=-1 &\text{ or } &3x=-2
\\&x=-1 &\text{ or } &x=-\frac{2}{3}
\end{array}$
Therefore the solution set is $\left\{-1, -\frac{2}{3}\right\}$.
