#### Answer

$\color{blue}{\left\{-2, 9\right\}}$

#### Work Step by Step

RECALL:
A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$.
The trinomial's factored form will be:
$x^2+bx+c=(x+d)(x+e)$
The trinomial in the equation has $b=-7$ and $c=-18$.
Note that $-18=-9(2)$ and $-7= -9+2$.
This means that $d=-9$ and $e=2$
Thus, the factored form of the trinomial is: $[w+(-9)](w+2)=(w-9)(w+2)$
The given equation maybe written as:
$(w-9)(w+2)=0$
Use the Zero-Product Property by equating each factor to zero.
Then, solve each equation to obtain:
\begin{array}{ccc}
&w-9 = 0 &\text{ or } &w+2=0
\\&w=9&\text{ or } &w=-2
\end{array}
Thus, the solution set is $\color{blue}{\left\{-2, 9\right\}}$.