2x^2-18x+80=-2x^2+18x

Solution for 2x^2-18x+80=-2x^2+18x equation:

2x^2-18x+80=-2x^2+18x

We move all terms to the left:

2x^2-18x+80-(-2x^2+18x)=0

We get rid of parentheses

2x^2+2x^2-18x-18x+80=0

We add all the numbers together, and all the variables

4x^2-36x+80=0

a = 4; b = -36; c = +80;
Δ = b2-4ac
Δ = -362-4·4·80
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4}{2*4}=\frac{32}{8}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4}{2*4}=\frac{40}{8}
=5
$