(12x^2-3)(x+3)+(2x^2+7x+3)(x-3)=0 giải pt ạ mng giúp em với

$(12x²-3)(x+3)+(2x²+7x+3)(x-3)=0$

$⇔12x³+36x²-3x-9+2x³-6x²+7x²-21x+3x-9=0$

$⇔14x³+37x²-21x-18=0$

$⇔(14x³+7x²)+(30x²+15x)-(36x+18)=0$

$⇔(2x+1)(7x²+15x-18)=0$

$⇔(2x+1)[(7x²+21x)-(6x+18)]=0$

$⇔(2x+1)(x+3)(7x-6)=0$

$⇔\left[ \begin{array}{l}2x+1=0\\x+3=0\\7x-6=0\end{array} \right.⇔\left[ \begin{array}{l}x=\frac{-1}{2}\\x=-3\\x=\frac{6}{7}\end{array} \right.$

Vậy $S=${$\frac{-1}{2};-3;\frac{6}{7}$}.