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

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

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

We multiply parentheses ..

(+8x^2-40x-16x+80)=0

We get rid of parentheses

8x^2-40x-16x+80=0

We add all the numbers together, and all the variables

8x^2-56x+80=0

a = 8; b = -56; c = +80;
Δ = b2-4ac
Δ = -562-4·8·80
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-24}{2*8}=\frac{32}{16}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+24}{2*8}=\frac{80}{16}
=5
$