n(n+1)=4032

Solution for n(n+1)=4032 equation:

n(n+1)=4032

We move all terms to the left:

n(n+1)-(4032)=0

We multiply parentheses

n^2+n-4032=0

a = 1; b = 1; c = -4032;
Δ = b2-4ac
Δ = 12-4·1·(-4032)
Δ = 16129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16129}=127$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-127}{2*1}=\frac{-128}{2}
=-64
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+127}{2*1}=\frac{126}{2}
=63
$