Answer
The solution set is $\left\{2, 5\right\}$.
Work Step by Step
Recall:
If the trinomial $x^2+bx+c$ is factorable, then its factored form is $(x+d)(x+e)$ where $c=de$ and $b=d+e$.
To check if the trinomial is factorable, look for factors of $10$ whose sum is equal to $-7$.
Note that $10=-5(-2)$ and $-7=-5+(-2)$.
Thus, $d=-5$ and $e=-2$, and the trinomial is factorable.
Hence,
$$p^2-7p+10=(p-5)(p-2)$$
The given equation is equivalent to $(p-5)(p-2)=0$.
Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain:
\begin{align*}
p-5&=0 &\text{or}& &p-2=0\\
p&=5 &\text{or}& &p=2\\
\end{align*}
Checking:
\begin{align*}
5^2-7(5)+10&=0\\
25-35+10&=0\\
0&\stackrel{\checkmark}=0
\end{align*}
\begin{align*}
2^2-7(2)+10&=0\\
4-14+10&=0\\
0&\stackrel{\checkmark}=0
\end{align*}
Therefore, the solution set is $\left\{2, 5\right\}$.
