(n-4)(n-2)=12

Solution for (n-4)(n-2)=12 equation:

(n-4)(n-2)=12

We move all terms to the left:

(n-4)(n-2)-(12)=0

We multiply parentheses ..

(+n^2-2n-4n+8)-12=0

We get rid of parentheses

n^2-2n-4n+8-12=0

We add all the numbers together, and all the variables

n^2-6n-4=0

a = 1; b = -6; c = -4;
Δ = b2-4ac
Δ = -62-4·1·(-4)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{13}}{2*1}=\frac{6-2\sqrt{13}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{13}}{2*1}=\frac{6+2\sqrt{13}}{2}
$