Answer
$x=\dfrac{1\pm\sqrt{47}i}{4}$
Work Step by Step
$2x^{2}-x+6=0$
Take the $6$ to the right side of the equation:
$2x^{2}-x=-6$
Take out common factor $2$ from the left side of the equation:
$2\Big(x^{2}-\dfrac{1}{2}x\Big)=-6$
Take the $2$ to divide the right side of the equation:
$x^{2}-\dfrac{1}{2}x=\dfrac{-6}{2}$
$x^{2}-\dfrac{1}{2}x=-3$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. For this case, $b=-\dfrac{1}{2}$:
$x^{2}-\dfrac{1}{2}x+\Big(\dfrac{-\frac{1}{2}}{2}\Big)^{2}=-3+\Big(\dfrac{-\frac{1}{2}}{2}\Big)^{2}$
$x^{2}-\dfrac{1}{2}x+\dfrac{1}{16}=-3+\dfrac{1}{16}$
$x^{2}-\dfrac{1}{2}x+\dfrac{1}{16}=-\dfrac{47}{16}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(x-\dfrac{1}{4}\Big)^{2}=-\dfrac{47}{16}$
Take the square root of both sides of the equation:
$\sqrt{\Big(x-\dfrac{1}{4}\Big)^{2}}=\sqrt{-\dfrac{47}{16}}$
$x-\dfrac{1}{4}=\pm\dfrac{\sqrt{47}}{4}i$
Solve for $x$:
$x=\dfrac{1\pm\sqrt{47}i}{4}$
