(n+4)(n-2)=36

Solution for (n+4)(n-2)=36 equation:

(n+4)(n-2)=36

We move all terms to the left:

(n+4)(n-2)-(36)=0

We multiply parentheses ..

(+n^2-2n+4n-8)-36=0

We get rid of parentheses

n^2-2n+4n-8-36=0

We add all the numbers together, and all the variables

n^2+2n-44=0

a = 1; b = 2; c = -44;
Δ = b2-4ac
Δ = 22-4·1·(-44)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{5}}{2*1}=\frac{-2-6\sqrt{5}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{5}}{2*1}=\frac{-2+6\sqrt{5}}{2}
$