Answer
$x=\dfrac{1\pm\sqrt{3}}{2}$
Work Step by Step
Using the properties of equality, the given equation, $
6x^2-3=6x
,$ is equivalent to
\begin{array}{l}\require{cancel}
6x^2-6x-3=0
\\\\
2x^2-2x-1=0
.\end{array}
Using completing the square, the solutions to the equation, $
2x^2-2x-1=0
,$ are
\begin{array}{l}\require{cancel}
x^2-x-\dfrac{1}{2}=0
\\\\
x^2-x=\dfrac{1}{2}
\\\\
x^2-x+\left( \dfrac{-1}{2} \right)^2=\dfrac{1}{2}+\left( \dfrac{-1}{2} \right)^2
\\\\
x^2-x+\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{4}
\\\\
x^2-x+\dfrac{1}{4}=\dfrac{3}{4}
\\\\
\left( x-\dfrac{1}{2} \right)^2=\dfrac{3}{4}
\\\\
x-\dfrac{1}{2}=\pm\sqrt{\dfrac{3}{4}}
\\\\
x-\dfrac{1}{2}=\pm\dfrac{\sqrt{3}}{2}
\\\\
x=\dfrac{1\pm\sqrt{3}}{2}
.\end{array}
