Answer
$p=-\displaystyle \frac{1}{2}+\frac{\sqrt{15}}{2}i$ or $p=-\displaystyle \frac{1}{2}-\frac{\sqrt{15}}{2}i$
Work Step by Step
$ p^{2}+p+4=0\qquad$... solve with the Quadractic formula. $a=1,\ b=1,\ c=4$
$ p=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $1, \ a$ for $1$ and $c$ for $4$.
$ p=\displaystyle \frac{-1\pm\sqrt{(1)^{2}-4\cdot(4)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$p=\displaystyle \frac{-1\pm\sqrt{1-16}}{2}$
$ p=\displaystyle \frac{-1\pm\sqrt{-15}}{2}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$)
$ p=\displaystyle \frac{-1\pm i\sqrt{15}}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$p=-\displaystyle \frac{1}{2}+\frac{\sqrt{15}}{2}i$ or $p=-\displaystyle \frac{1}{2}-\frac{\sqrt{15}}{2}i$
