## Intermediate Algebra: Connecting Concepts through Application

$\left\{-\dfrac{5}{4}, \dfrac{8}{7}\right\}$
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*}