Answer
$\displaystyle \frac{1}{2}\pm\frac{\sqrt{3}}{2}i$
Work Step by Step
Use the quadratic formula
$a=1, b=-1, c=1$
$x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$x=\displaystyle \frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\sqrt{-3}}{2}$
...If $N$ is a positive real number, we define the principal square root of $-N,$ denoted by $\sqrt{-N},$ as $\sqrt{-N}=\sqrt{N}i$
$=\displaystyle \frac{1\pm i\sqrt{3}}{2}$
$=\displaystyle \frac{1}{2}\pm\frac{\sqrt{3}}{2}i$
