n(n+1)=6480

Solution for n(n+1)=6480 equation:

n(n+1)=6480

We move all terms to the left:

n(n+1)-(6480)=0

We multiply parentheses

n^2+n-6480=0

a = 1; b = 1; c = -6480;
Δ = b2-4ac
Δ = 12-4·1·(-6480)
Δ = 25921
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25921}=161$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-161}{2*1}=\frac{-162}{2}
=-81
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+161}{2*1}=\frac{160}{2}
=80
$