The square of the solution of the equation... The square of the solution of the equation 
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
is equal to: ...
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
$\implies x\sqrt{7} = \sqrt{8 + 3\sqrt{7}} - \sqrt{8 - 3\sqrt{7}}$
$\implies 7x^2 = \left(8 + 3\sqrt{7}\right) + \left(8 - 3\sqrt{7}\right) - \underline{2\sqrt{\left(8+3\sqrt{7}\right)\cdot\left(8-3\sqrt{7}\right)}}$
$\implies 7x^2 = 16 - 2\cdot\sqrt{64-63} = 14$
$\implies x^2 = 2$
I know how to do all the steps up to the part where the underlined section comes into play, can someone please explain where does this come from. Thanks! 

The formula $(a-b)^2=a^2+b^2-2ab$  is used.
Here, $a=\sqrt{8+3\sqrt7}$ and $b=\sqrt{8-3\sqrt7}$.
