\(2x^2-6x+1=0\\ \Leftrightarrow x^2-3x+\dfrac{1}{2}=0\\ \Leftrightarrow x^2-3x+\dfrac{9}{4}-\dfrac{7}{4}=0\\ \Leftrightarrow\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{7}{4}=0\\ \Leftrightarrow\left(x-\dfrac{3}{2}\right)^2-\dfrac{7}{4}=0\\ \Leftrightarrow\left(x-\dfrac{3}{2}+\dfrac{\sqrt{7}}{2}\right)\left(x-\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}\right)=0\\ \Leftrightarrow\left(x-\dfrac{3-\sqrt{7}}{2}\right)\left(x-\dfrac{3+\sqrt{7}}{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3-\sqrt{7}}{2}=0\\x-\dfrac{3+\sqrt{7}}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{7}}{2}\\x=\dfrac{3+\sqrt{7}}{2}\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{3+\sqrt{7}}{2};\dfrac{3-\sqrt{7}}{2}\right\}\)