Answer
$\left\{-\dfrac{5}{4}, \dfrac{8}{7}\right\}$
Work Step by Step
Write the equation in the form $ax^2+bx+c=0$ to obtain:
\begin{align*}
28p^2+3p+60-100&=0\\
28p^2+3p-40&=0
\end{align*}
Solve the equation using the $\text{ac-method}$.
With $a=28$ and $c=-40$, $ac= 28(-40)=-1120$.
Recall:
$ax^2+bx+c$ can be factored if there exists $d$ and $e$ such that $ac=de$ and $d+e=b$.
The trinomial can then be tfactored by grouping by rewriting the trinomial as $ax^2+dx+ex+c$.
Look for factors of $-1120$ whose sum is $3$.
Note that $-1120=-32(35)$ and $-32+35=3$.
This means that the trinomial is factorable and $d=-32$ and $e=35$.
Solve the given equation by grouping:
\begin{align*}
28p^2+3p-40&=0\\
28p^2-32p+35p-40&=0\\
(28p^2-32p)+(35p-40)&=0\\
4p(7p-8)+5(7p-8)&=0\\
(7p-8)(4p+5)&=0\
\end{align*}
Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain:
\begin{align*}
7p-8&=0 &\text{or}& &4p+5=0\\
7p&=8 &\text{or}& &4p=-5\\
p&=\frac{8}{7} &\text{or}& &p=-\frac{5}{4}
\end{align*}
