x^2-4X=96^2

Solution for x^2-4X=96^2 equation:

x^2-4x=96^2

We move all terms to the left:

x^2-4x-(96^2)=0

We add all the numbers together, and all the variables

x^2-4x-9216=0

a = 1; b = -4; c = -9216;
Δ = b2-4ac
Δ = -42-4·1·(-9216)
Δ = 36880
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{36880}=\sqrt{16*2305}=\sqrt{16}*\sqrt{2305}=4\sqrt{2305}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{2305}}{2*1}=\frac{4-4\sqrt{2305}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{2305}}{2*1}=\frac{4+4\sqrt{2305}}{2}
$