2x^2-8x=4x+80

Solution for 2x^2-8x=4x+80 equation:

2x^2-8x=4x+80

We move all terms to the left:

2x^2-8x-(4x+80)=0

We get rid of parentheses

2x^2-8x-4x-80=0

We add all the numbers together, and all the variables

2x^2-12x-80=0

a = 2; b = -12; c = -80;
Δ = b2-4ac
Δ = -122-4·2·(-80)
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{-(-12)-28}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+28}{2*2}=\frac{40}{4}
=10
$