Answer
The solutions are $x=\dfrac{1}{4}\pm\dfrac{\sqrt{19}}{4}i$
Work Step by Step
$4x^{2}-2x+5=0$
Take $5$ to the right side of the equation:
$4x^{2}-2x=-5$
Take out common factor $4$ from the left side:
$4\Big(x^{2}-\dfrac{1}{2}x\Big)=-5$
Take $4$ to divide the right side:
$x^{2}-\dfrac{1}{2}x=-\dfrac{5}{4}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this particular case, $b=-\dfrac{1}{2}$:
$x^{2}-\dfrac{1}{2}x+\Big(-\dfrac{1}{2\cdot2}\Big)^{2}=-\dfrac{5}{4}+\Big(-\dfrac{1}{2\cdot2}\Big)^{2}$
$x^{2}-\dfrac{1}{2}x+\dfrac{1}{16}=-\dfrac{5}{4}+\dfrac{1}{16}$
$x^{2}-\dfrac{1}{2}x+\dfrac{1}{16}=-\dfrac{19}{16}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(x-\dfrac{1}{4}\Big)^{2}=-\dfrac{19}{16}$
Take the square root of both sides:
$\sqrt{\Big(x-\dfrac{1}{4}\Big)^{2}}=\pm\sqrt{-\dfrac{19}{16}}$
$x-\dfrac{1}{4}=\pm\dfrac{\sqrt{19}}{4}i$
Solve for $x$:
$x=\dfrac{1}{4}\pm\dfrac{\sqrt{19}}{4}i$
