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