(12x²-3)(x+3)+(2x²+7x+3)(x-3)=0
⇔(12x²-3)(x+3)+(2x²+x+6x+3)(x-3)=0
⇔(12x²-3)(x+3)+[x(2x+1)+3.(2x+1)(x-3)=0
⇔(12x²-3)(x+3)+(2x+1)(x+3)(x-3)=0
⇔(x+3)[12x²-3+(2x+1)(x-3)]=0
⇔(x+3)[3.(4x²-1)+(2x+1)(x-3)]=0
⇔(x+3)[3.(2x-1)(2x+1)+(2x+1)(x-3)]=0
⇔(x+3)(2x+1)(6x-3+x-3)=0
⇔(x+3)(2x+1)(7x-6)=0
⇔x+3=0; 2x+1=0; 7x-6=0
⇔x=-3; x=-1/2; x=6/7
Vậy S={-3;-1/2;6/7}