$2(9x^2+6x+1)=(3x+1)(x-2)$
⇔ $2(3x+1)^2-(3x+1)(x-2)=0$
⇔ $(3x+1)[2(3x+1)-(x-2)]=0$
⇔ $(3x+1)(6x+2-x+2)=0$
⇔ $(3x+1)(5x+4)=0$
⇔ \(\left[ \begin{array}{l}3x+1=0\\5x+4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}3x=-1\\5x=-4\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-1}{3}\\x=\frac{-4}{5}\end{array} \right.\)
Vậy $S=${$\frac{-1}{3};\frac{-4}{5}$}
