$(2x^2+1)(4x-3)=(2x^2+1)(x-12)$
⇔ $(2x^2+1)(4x-3)-(2x^2+1)(x-12)=0$
⇔ $(2x^2+1)(4x-3-x+12)=0$
⇔ $(2x^2+1)(3x+9)=0$
⇔ \(\left[ \begin{array}{l}2x^2+1=0\\3x+9=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x^2=-1\\3x=-9\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x^2=\frac{-1}{2}(vô lí)\\x=-3\end{array} \right.\)
Vậy $S=${$-3$}