Answer
$\frac{1}{2}\pm\frac{1}{2}i$
Work Step by Step
Given: $2x^{2}-2x+1=0$
We solve using the quadratic formula
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here, $a=2, b=-2, c=1$
$x=\displaystyle \frac{-(-2)\pm\sqrt{(-2)^{2}-4(2)(1)}}{2(2)}$
$=\frac{2\pm\sqrt{4-8}}{4}$
$=\frac{2\pm\sqrt{-4}}{4}$
$=\frac{2\pm\sqrt{-1*4}}{4}$
We use the fact that $\sqrt{-1}=i$:
$=\frac{2\pm 2i}{4}$
$=\frac{1}{2}\pm\frac{1}{2}i$
