Answer
$x=\left\{ \dfrac{5-\sqrt{105}}{20},\dfrac{5+\sqrt{105}}{20} \right\}$
Work Step by Step
Using the completing the square method, the solutions of the given quadratic equation, $
2x^2-x+6=0
,$ is
\begin{array}{l}\require{cancel}
\dfrac{1}{2}\cdot(2x^2-x+6)=(0)\cdot\dfrac{1}{2}
\\\\
x^2-\dfrac{1}{2}x+3=0
\\\\
x^2-\dfrac{1}{2}x=-3
\\\\
x^2-\dfrac{1}{2}x+\left( \dfrac{-1}{2\cdot2} \right)^2=\dfrac{1}{5}+\left( \dfrac{-1}{2\cdot2} \right)^2
\\\\
x^2-\dfrac{1}{2}x+\dfrac{1}{16}=\dfrac{1}{5}+\dfrac{1}{16}
\\\\
\left( x-\dfrac{1}{4}\right)^2=\dfrac{16}{80}+\dfrac{5}{80}
\\\\
\left( x-\dfrac{1}{4}\right)^2=\dfrac{21}{80}
\\\\
x-\dfrac{1}{4}=\pm\sqrt{\dfrac{21}{80}}
\\\\
x-\dfrac{1}{4}=\pm\sqrt{\dfrac{21}{80}\cdot\dfrac{5}{5}}
\\\\
x-\dfrac{1}{4}=\pm\sqrt{\dfrac{105}{400}}
\\\\
x-\dfrac{1}{4}=\pm\dfrac{\sqrt{105}}{20}
\\\\
x=\dfrac{1}{4}\pm\dfrac{\sqrt{105}}{20}
\\\\
x=\dfrac{5}{20}\pm\dfrac{\sqrt{105}}{20}
\\\\
x=\dfrac{5\pm\sqrt{105}}{20}
.\end{array}
Hence, $
x=\left\{ \dfrac{5-\sqrt{105}}{20},\dfrac{5+\sqrt{105}}{20} \right\}
.$
