Solve each of the following quadratic equations:

We write, $-(2 b-1) x=-(b-5) x-(b+4) x$ as $x^{2} \times\left(b^{2}-b-20\right)=\left(b^{2}-b-20\right) x^{2}=[-(b-5) x] \times[-(b+4) x]$

$\therefore x^{2}-(2 b-1) x+\left(b^{2}-b-20\right)=0$

$\Rightarrow x^{2}-(b-5) x-(b+4) x+(b-5)(b+4)=0$

$\Rightarrow x[x-(b-5)]-(b+4)[x-(b-5)]=0$

$\Rightarrow[x-(b-5)][x-(b+4)]=0$

$\Rightarrow x-(b-5)=0$ or $x-(b+4)=0$

$\Rightarrow x=b-5$ or $x=b+4$

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