Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

The given equation is $x^{2}+5 x-\left(a^{2}+a-6\right)=0$.

Comparing it with $A x^{2}+B x+C=0$, we get

$A=1, B=5$ and $C=-\left(a^{2}+a-6\right)$

$\therefore$ Discriminant, $D=B^{2}-4 A C=5^{2}-4 \times 1 \times\left[-\left(a^{2}+a-6\right)\right]=25+4 a^{2}+4 a-24=4 a^{2}+4 a+1=(2 a+1)^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{(2 a+1)^{2}}=2 a+1$

$\therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-5+2 a+1}{2 \times 1}=\frac{2 a-4}{2}=a-2$

$\beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-5-(2 a+1)}{2 \times 1}=\frac{-2 a-6}{2}=-a-3=-(a+3)$

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