Solve each of the following quadratic equations:

We write, $5 x=(a+3) x-(a-2) x$ as $x^{2} \times\left[-\left(a^{2}+a-6\right)\right]=-\left(a^{2}+a-6\right) x^{2}=(a+3) x \times[-(a-2) x]$

$\therefore x^{2}+5 x-\left(a^{2}+a-6\right)=0$

$\Rightarrow x^{2}+(a+3) x-(a-2) x-(a+3)(a-2)=0$

$\Rightarrow x[x+(a+3)]-(a-2)[x+(a+3)]=0$

$\Rightarrow[x+(a+3)][x-(a-2)]=0$

$\Rightarrow x+(a+3)=0$ or $x-(a-2)=0$

$\Rightarrow x=-(a+3)$ or $x=a-2$

Hence, $-(a+3)$ and $(a-2)$ are the roots of the given equation.