$( x-2 ).(3+x)=2. (x-2)$
⇔ $3x+x^2-6-2x=2x-4$
⇔ $x^2+x-6-2x+4=0$
⇔ $x^2-x-2=0$
⇔ $x^2-2x+x-2=0$
⇔ $x(x-2)+(x-2)=0$
⇔ $(x-2)(x+1)=0$
⇔ \(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Vậy $S=${$-1;2$}