2x^2=96+4x

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

2x^2=96+4x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2-4x-96=0

a = 2; b = -4; c = -96;
Δ = b2-4ac
Δ = -42-4·2·(-96)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-28}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+28}{2*2}=\frac{32}{4}
=8
$