2x^2-4x=x^2+96

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

2x^2-4x=x^2+96

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-4x-96=0

a = 1; b = -4; c = -96;
Δ = b2-4ac
Δ = -42-4·1·(-96)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*1}=\frac{24}{2}
=12
$