4x^2+80=-36x

Solution for 4x^2+80=-36x equation:

4x^2+80=-36x

We move all terms to the left:

4x^2+80-(-36x)=0

We get rid of parentheses

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