0=4x^2+20x

Solution for 0=4x^2+20x equation:

0=4x^2+20x

We move all terms to the left:

0-(4x^2+20x)=0

We add all the numbers together, and all the variables

-(4x^2+20x)=0

We get rid of parentheses

-4x^2-20x=0

a = -4; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·(-4)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*-4}=\frac{0}{-8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*-4}=\frac{40}{-8}
=-5
$