\(2x^2-6x+1=0\)
\(\Leftrightarrow x^2-3x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x^2-3x+\left(\dfrac{3}{2}\right)^2=\dfrac{7}{4}\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\left(\pm\dfrac{\sqrt{7}}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{7}}{2}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{7}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{7}+3}{2}\\x=\dfrac{-\sqrt{7}+3}{2}\end{matrix}\right.\)
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