Answer
The solution set is $\left\{-9, 1\right\}$.
Work Step by Step
Recall:
If $x^2+bx+c$ is factorable, then its factored forom is $(x+d)(x+e)$ where $c=de$ and $b=d+e$.
In the given equation, the quadratic trinomial has $c=-9$ and $b=8$.
Note that $-9=-1(9)$ and $8=-1+9 $.
This means that $d=-1$ and $e=9$.
Hence, factoring the trinomial gives:
$$(r-1)(r+9)=0$$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
\begin{align*}
r-1&=0 &\text{or}& &r+9=0\\
r&=1 &\text{or}& &r=-9\\
\end{align*}
Thus, the solution set is $\left\{-9, 1\right\}$.
