Question:

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

Solution:

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

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

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

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

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

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

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

$\Rightarrow x=-\frac{a+b}{2}$ or $x=\frac{a-b}{2}$

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