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

The given equation is $x^{2}-(2 b-1) x+\left(b^{2}-b-20\right)=0$.

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

$A=1, B=-(2 b-1)$ and $C=b^{2}-b-20$

$\therefore$ Discriminant, $D=B^{2}-4 A C=[-(2 b-1)]^{2}-4 \times 1 \times\left(b^{2}-b-20\right)=4 b^{2}-4 b+1-4 b^{2}+4 b+80=81>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{81}=9$

$\therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-[-(2 b-1)]+9}{2 \times 1}=\frac{2 b+8}{2}=b+4$

$\beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-[-(2 b-1)]-9}{2 \times 1}=\frac{2 b-10}{2}=b-5$

Hence, $(b+4)$ and $(b-5)$ are the roots of the given equation.