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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(4(-4x^2+20x)), so:

4(-4x^2+20x)

We multiply parentheses

-16x^2+80x

Back to the equation:

-(-16x^2+80x)


We get rid of parentheses

16x^2-80x=0

a = 16; b = -80; c = 0;
Δ = b2-4ac
Δ = -802-4·16·0
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-80}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+80}{2*16}=\frac{160}{32}
=5
$