We write, $-(2 b-1) x=-(b-5) x-(b+4) x$ as
$\begin{array}{l}
x^{2} \times\left(b^{2}-b-20\right)=\left(b^{2}-b-20\right) x^{2}=[-(b-5) x] \times[-(b+4) x] \\
\therefore x^{2}-(2 b-1) x+\left(b^{2}-b-20\right)=0 \\
\Rightarrow x^{2}-(b-5) x-(b+4) x+(b-5)(b+4)=0 \\
\Rightarrow x[x-(b-5)]-(b+4)[x-(b-5)]=0 \\
\Rightarrow[x-b-5][x-(b+4)]=0 \\
\Rightarrow x-(b-5)=0 \text { or } x-(b+4)=0 \\
\Rightarrow x=b-5 \text { or } x=b+4
\end{array}$

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