We write, $5 x=(a+3) x-(a-2) x$ as

$\begin{array}{l}
x^{2} \times\left[-\left(a^{2}+a-6\right)\right]=-\left(a^{2}+a-6\right) x^{2}=(a+3) x \times[-(a-2) x] \\
\therefore x^{2}+5 x-\left(a^{2}+a-6\right)=0 \\
\Rightarrow x^{2}+(a+3) x-(a-2) x-(a+3)(a-2)=0 \\
\Rightarrow x[x+(a+3)]-(a-2)[x+(a+3)]=0 \\
\Rightarrow[x+(a+3)][x-(a-2)]=0 \\
\Rightarrow x+(a+3)=0 x-(a-2)=0 \\
\Rightarrow x=-(a+3) \text { or } x=a-2
\end{array}$

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