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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2-6x-88=0

a = 4; b = -6; c = -88;
Δ = b2-4ac
Δ = -62-4·4·(-88)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-38}{2*4}=\frac{-32}{8}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+38}{2*4}=\frac{44}{8}
=5+1/2
$