(n+2)(n-4)=24

Solution for (n+2)(n-4)=24 equation:

(n+2)(n-4)=24

We move all terms to the left:

(n+2)(n-4)-(24)=0

We multiply parentheses ..

(+n^2-4n+2n-8)-24=0

We get rid of parentheses

n^2-4n+2n-8-24=0

We add all the numbers together, and all the variables

n^2-2n-32=0

a = 1; b = -2; c = -32;
Δ = b2-4ac
Δ = -22-4·1·(-32)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{33}}{2*1}=\frac{2-2\sqrt{33}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{33}}{2*1}=\frac{2+2\sqrt{33}}{2}
$