\(6x^2-x-12=0\)
\(\Leftrightarrow6x^2-9x+8x-12=0\)
\(\Leftrightarrow3x\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\3x+4=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(6x^2-x-12=0\) (1)
\(\Leftrightarrow6x^2+8x-9x-12=0\)
\(\Leftrightarrow2x\left(3x+4\right)-3\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-4\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-\dfrac{4}{3};\dfrac{3}{2}\right\}\)