Answer
$p=\displaystyle \frac{1}{2}+\frac{\sqrt{3}}{2}i$ or $p=\displaystyle \frac{1}{2}-\frac{\sqrt{3}}{2}i$
Work Step by Step
$ p^{2}-p+1=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-1,\ c=1$
$ p=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-1,\ a$ for $1$ and $c$ for $1$.
$ p=\displaystyle \frac{-(-1)\pm\sqrt{(-1)^{2}-4\cdot(1)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$p=\displaystyle \frac{1\pm\sqrt{1-4}}{2}$
$ p=\displaystyle \frac{1\pm\sqrt{-3}}{2}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$)
$ p=\displaystyle \frac{1\pm i\sqrt{3}}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$p=\displaystyle \frac{1}{2}+\frac{\sqrt{3}}{2}i$ or $p=\displaystyle \frac{1}{2}-\frac{\sqrt{3}}{2}i$
