(4n-2)(2n-4)=0

Solution for (4n-2)(2n-4)=0 equation:

(4n-2)(2n-4)=0

We multiply parentheses ..

(+8n^2-16n-4n+8)=0

We get rid of parentheses

8n^2-16n-4n+8=0

We add all the numbers together, and all the variables

8n^2-20n+8=0

a = 8; b = -20; c = +8;
Δ = b2-4ac
Δ = -202-4·8·8
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12}{2*8}=\frac{8}{16}
=1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12}{2*8}=\frac{32}{16}
=2
$