a) $x^2 + 2x - 3x - 6 = 0$
⇔ $(x^2 + 2x )-( 3x + 6) = 0$
⇔ $ x(x+2) -3(x+2)=0$
⇔$ (x+2)(x-3)=0$
⇔\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
b) $x^2 +3x - 2x - 6 =0$
⇔ $(x^2 +3x )- (2x + 6)=0$
⇔$x (x+3) -2 (x+3) =0$
⇔$(x+3)(x-2)=0$
⇔ \(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\)