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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

n^2-4n-32=0

a = 1; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·1·(-32)
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{-(-4)-12}{2*1}=\frac{-8}{2}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*1}=\frac{16}{2}
=8
$