We have, $-2 a x=(2 b-a) x-(2 b+a) x$ as

$\begin{array}{l}
x^{2} \times\left[-\left(4 b^{2}-a^{2}\right)\right]=-\left(4 b^{2}-a^{2}\right) x^{2}=(2 b-a) x \times[-(2 b+a) x] \\
\therefore x^{2}-2 a x-\left(4 b^{2}-a^{2}\right)=0 \\
\Rightarrow x^{2}+(2 b-a) x-(2 b+a) x-(2 b-a)(2 b+a)=0 \\
\Rightarrow x[x+(2 b-a)]-(2 b+a)[x+(2 b-a)]=0 \\
\Rightarrow[x+(2 b-a)][x-(2 b+a)]=0 \\
\Rightarrow x+(2 b-a)=0 \text { or } x-(2 b+a)=0 \\
x=-(2 b-a) \text { or } x=2 b+a \\
\Rightarrow x=a-2 b \text { or } x=a+2 b
\end{array}$

Hence, $a-2 b$ and $a+2 b$ are the roots of the given equation.