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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2-6n+4=0

a = 2; b = -6; c = +4;
Δ = b2-4ac
Δ = -62-4·2·4
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2}{2*2}=\frac{4}{4}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2}{2*2}=\frac{8}{4}
=2
$