Answer
$p=\frac{4}{3}$ or $x=-18$
Work Step by Step
$ 72=3p^{2}+50p\qquad$...add $(-3p^{2}-50p)$ to both sides.
$ 72-3p^{2}-50p=3p^{2}+50p-3p^{2}-50p\qquad$...add like terms.
$-3p^{2}-50p+72=0\qquad$...mulitiply both sides by $-1$.
$ 3p^{2}+50p-72=0\qquad$... solve with the Quadractic formula. $a=3,\ b=50,\ c=-72$
$ p=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $50,\ a$ for $3$ and $c$ for $-72$.
$ p=\displaystyle \frac{-50\pm\sqrt{(50)^{2}-4\cdot(-72)\cdot 3}}{2\cdot 3}\qquad$... simplify.
$p=\displaystyle \frac{-50\pm\sqrt{2500+864}}{6}$
$p=\displaystyle \frac{-50\pm\sqrt{3364}}{6}$
$ p=\displaystyle \frac{-50\pm 58}{6}\qquad$... the symbol $\pm$ indicates two solutions.
$p=\displaystyle \frac{-50+58}{6}=\frac{8}{6}=\frac{4}{3}$ or $x=\displaystyle \frac{-50-58}{6}=\frac{-108}{6}=-18$
