$(x-1)(5x+3)=(3x-8)(x-1)$
$⇔(x-1)(5x+3-3x+8)=0$
$⇔(x-1)(2x+11)=0$
$⇔$ \(\left[ \begin{array}{l}x-1=0\\2x+11=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x-1=0\\2x=-11\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=1\\x=\frac{-11}{2}\end{array} \right.\)
$Vậy...$