(2x-8)(2x-4)=0

Solution for (2x-8)(2x-4)=0 equation:

(2x-8)(2x-4)=0

We multiply parentheses ..

(+4x^2-8x-16x+32)=0

We get rid of parentheses

4x^2-8x-16x+32=0

We add all the numbers together, and all the variables

4x^2-24x+32=0

a = 4; b = -24; c = +32;
Δ = b2-4ac
Δ = -242-4·4·32
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8}{2*4}=\frac{16}{8}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8}{2*4}=\frac{32}{8}
=4
$