Answer
The solution set is $\left\{-\frac{1}{2}, 3\right\}$.
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 $a=2$, $b=-5$, and $c=-3$.
Thus, $ac = 2(-3) = -6$
The factors of $-6$ whose sum is equal to the coefficient of the middle term (which is $b=-5$) are $-6$ and $1$.
This means that $d=-6$ and $e=1$.
Rewrite the middle term of the trinomial as $-6x$ +$x$ to obtain:
$2x^2-5x-3 = 0
\\2x^2+(-6x)+x -3 = 0$
Regroup then factor out the GCF in each group to obtain:
$(2x^2-6x)+(x-3)=0
\\2x(x-3) +(1)(x-3)=0$
Factor out $x-3$ to obtain:
$(x-3)(2x+1)=0$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
$\begin{array}{ccc}
&x-3=0 &\text{ or } &2x+1=0
\\&x=3 &\text{ or } &2x=-1
\\&x=3 &\text{ or } &x=-\frac{1}{2}
\end{array}$
Therefore the solution set is $\left\{-\frac{1}{2}, 3\right\}$.
