Answer
$x=\left\{ \dfrac{2-i\sqrt{2}}{2},\dfrac{2+i\sqrt{2}}{2} \right\}$
Work Step by Step
Using the completing the square method, the solutions of the given quadratic equation, $
2x^2-4x=-3
,$ is
\begin{array}{l}\require{cancel}
\dfrac{1}{2}\cdot (2x^2-4x)=(-3)\cdot\dfrac{1}{2}
\\\\
x^2-2x=-\dfrac{3}{2}
\\\\
x^2-2x+\left( \dfrac{-2}{2} \right)^2=-\dfrac{3}{2}+\left( \dfrac{-2}{2} \right)^2
\\\\
x^2-2x+1=-\dfrac{3}{2}+1
\\\\
x^2-2x+1=-\dfrac{3}{2}+\dfrac{2}{2}
\\\\
(x-1)^2=-\dfrac{1}{2}
\\\\
x-1=\pm\sqrt{-\dfrac{1}{2}}
\\\\
x-1=\pm i\sqrt{\dfrac{1}{2}}
\\\\
x-1=\pm i\sqrt{\dfrac{1}{2}\cdot\dfrac{2}{2}}
\\\\
x-1=\pm i\sqrt{\dfrac{2}{4}}
\\\\
x-1=\pm i\dfrac{\sqrt{2}}{2}
\\\\
x=1\pm i\dfrac{\sqrt{2}}{2}
\\\\
x=\dfrac{2}{2}\pm i\dfrac{\sqrt{2}}{2}
\\\\
x=\dfrac{2\pm i\sqrt{2}}{2}
.\end{array}
Hence, $
x=\left\{ \dfrac{2-i\sqrt{2}}{2},\dfrac{2+i\sqrt{2}}{2} \right\}
.$
