Solve the following quadratic equation for $x:4x^{2} +\ 4bx\ –\ ( a^{2} \ –\ b^{2}) =\ 0$

Given: The equation:$4x^{2} +\ 4bx\ –\ \left( a^{2} \ –\ b^{2}\right) =\ 0$

To do: To find the value of $x$.

The given quadratic equation $4x^{2} +\ 4bx\ –\ \left( a^{2} \ –\ b^{2}\right)=0$

On dividing the given equation by 4,

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

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

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

$\Rightarrow x^{2} +2\left(\frac{b}{2}\right) x+\left(\frac{b}{2}\right)^{2} =\left(\frac{a^{2} -b^{2}}{4}\right) +\left(\frac{b}{2}\right)^{2} \ \ \ \ \ \left( Let\ us\ add\ \left(\frac{b}{2}\right)^{2} to\ both\ the\ sides\right)$

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

$\Rightarrow \left( x+\frac{b}{2}\right)^{2} =\left(\frac{a}{2}\right)^{2}$

$\Rightarrow \left( x+\frac{b}{2}\right) =\pm \left(\frac{a}{2}\right)$

If $\left( x+\frac{b}{2}\right) =\left(\frac{a}{2}\right)$

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

If $\left( x+\frac{b}{2}\right) =-\left(\frac{a}{2}\right)$

$\Rightarrow x=-\frac{a}{2} -\frac{b}{2} =-\left(\frac{a-b}{2}\right)$

Therefore the given quadratic equation has two solution $x=\frac{a-b}{2}$  and $x=-( \frac{a-b}{2})$