#### Answer

$\color{blue}{\left\{5, -\frac{3}{2}\right\}}$

#### Work Step by Step

Add $5$ to both sides to obtain:
$2x^2+13x+15=0$
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 above has $a=2$, $b=13$, and $c=15$.
Thus, $ac = 2(15) = 30$
Note that $30=10(3)$ and $10+3=13$
This means that $d=10$ and $e=3$.
Rewrite the middle term of the trinomial as $10x$ +$3x$ to obtain:
$2x^2+13x+15 =2x^2+10x+3x+15 $
Group the first two terms together and the last two terms together.
Then, factor out the GCF in each group to obtain:
$=(2x^2+10x)+(3x+15)
\\=2x(x+5) +3(x+5)$
Factor out the GCF $x+5$ to obtain:
$=(x+5)(2x+3)$
This means that the equation above can be written as:
$(x+5)(2x+3)=0$
Use the Zero-Factor Property by equating each factor to zero. Then, solve each equation to obtain:
\begin{array}{ccc}
&x+5=0 &\text{or} &2x+3=0
\\&x=-5 &\text{or} &2x=-3
\\&x=-5 &\text{or} & x=-\frac{3}{2}
\end{array}
Checking:
If x=-5:
$2(-5)^2+13(-5)+10=-5?
\\2(25)+(-65)+10=-5?
\\50-65+10=-5?
\\-5=-5$
If $x=-\frac{3}{2}$:
$2(-\frac{3}{2})^2+13(-\frac{3}{2})+10=-5?
\\2(\frac{9}{4}-\frac{39}{2}+10=-5?
\\\frac{9}{2} -\frac{39}{2} + \frac{20}{2} =-5?
\\\frac{-10}{2}=-5?
\\-5=-5$
Thus, the solution set is $\color{blue}{\left\{5, -\frac{3}{2}\right\}}$.