Solve each of the following quadratic equations:

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

$\therefore x^{2}-4 a x-b^{2}+4 a^{2}=0$

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

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

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

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

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

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

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