Answer
$p=1$ and $p=-12$
Work Step by Step
$p^{2}+11p-12=0$
Use the quadratic formula to solve this equation. The formula is $p=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
For this equation, $a=1$, $b=11$ and $c=-12$
Substitute:
$p=\dfrac{-11\pm\sqrt{11^{2}-4(1)(-12)}}{2(1)}=\dfrac{-11\pm\sqrt{121+48}}{2}=...$
$...=\dfrac{-11\pm\sqrt{169}}{2}=\dfrac{-11\pm13}{2}$
Our two solutions are:
$p=\dfrac{-11+13}{2}=1$
$p=\dfrac{-11-13}{2}=-12$
