\(6y^2-5y-38=0\\ \Leftrightarrow y^2-\dfrac{5}{6}y-\dfrac{38}{6}=0\\ \Leftrightarrow y^2-\dfrac{5}{6}y+\dfrac{25}{144}-\dfrac{937}{144}=0\\ \Leftrightarrow\left(y^2-\dfrac{5}{6}y+\dfrac{25}{144}\right)-\dfrac{937}{144}=0\\ \Leftrightarrow\left(y-\dfrac{5}{12}\right)^2-\dfrac{937}{144}=0\\ \Leftrightarrow\left(y-\dfrac{5}{12}+\dfrac{\sqrt{937}}{12}\right)\left(y-\dfrac{5}{12}-\dfrac{\sqrt{937}}{12}\right)=0\\ \Leftrightarrow\left(y-\dfrac{5-\sqrt{937}}{12}\right)\left(y-\dfrac{5+\sqrt{937}}{12}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y-\dfrac{5-\sqrt{937}}{12}=0\\y-\dfrac{5+\sqrt{937}}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{5-\sqrt{937}}{12}\\y=\dfrac{5+\sqrt{937}}{12}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{5-\sqrt{937}}{12};\dfrac{5+\sqrt{937}}{12}\right\}\)
