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

The given equation is $x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$.

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

$A=1, B=6$ and $C=-\left(a^{2}+2 a-8\right)$

$\therefore$ Discriminant, $D=B^{2}-4 A C=6^{2}-4 \times 1 \times\left[-\left(a^{2}+2 a-8\right)\right]=36+4 a^{2}+8 a-32=4 a^{2}+8 a+4=4\left(a^{2}+2 a+1\right)=4(a+1)^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{4(a+1)^{2}}=2(a+1)$

$\therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-6+2(a+1)}{2 \times 1}=\frac{2 a-4}{2}=a-2$

$\beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-6-2(a+1)}{2 \times 1}=\frac{-2 a-8}{2}=-a-4=-(a+4)$

Hence, $(a-2)$ and $-(a+4)$ are the roots of the given equation.