Lời giải:
Ta có:
$(x-1).(5x+3)=(3x-8).(x-1)$
$=(x-1).(5x+3)-(3x-8).(x-1)=0$
$=(x-1).(5x+3-3x+8)=0$
$=(x-1).(2x+11)=0$
<=>\(\left[ \begin{array}{l}x-1=0\\2x+11=0\end{array} \right.\)
<=>$\left \{ {{x=1} \atop {x=\frac{-11}{2}}} \right.$
Vậy $S=${$\frac{-11}{2};1$}