Solve each of the following quadratic equations:

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

$\therefore x^{2}-2 a x-\left(4 b^{2}-a^{2}\right)=0$

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

$\Rightarrow x[x+(2 b-a)]-(2 b+a)[x+(2 b-a)]=0$

$\Rightarrow[x+(2 b-a)][x-(2 b+a)]=0$

$\Rightarrow x+(2 b-a)=0$ or $x-(2 b+a)=0$

$\Rightarrow x=-(2 b-a)$ or $x=2 b+a$

$\Rightarrow x=a-2 b$ or $x=a+2 b$

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