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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

n^2-6n+8=0

a = 1; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·1·8
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*1}=\frac{4}{2}
=2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2}{2*1}=\frac{8}{2}
=4
$