n(n+4)-4n=36

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

n(n+4)-4n=36

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We add all the numbers together, and all the variables

n^2-36=0

a = 1; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·1·(-36)
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{-(0)-12}{2*1}=\frac{-12}{2}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*1}=\frac{12}{2}
=6
$