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

Solution for (n+-4)(n+-8)=0 equation:

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

We add all the numbers together, and all the variables

(n-4)(n-8)=0

We multiply parentheses ..

(+n^2-8n-4n+32)=0

We get rid of parentheses

n^2-8n-4n+32=0

We add all the numbers together, and all the variables

n^2-12n+32=0

a = 1; b = -12; c = +32;
Δ = b2-4ac
Δ = -122-4·1·32
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*1}=\frac{8}{2}
=4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*1}=\frac{16}{2}
=8
$