n(n+1)=399

Solution for n(n+1)=399 equation:

n(n+1)=399

We move all terms to the left:

n(n+1)-(399)=0

We multiply parentheses

n^2+n-399=0

a = 1; b = 1; c = -399;
Δ = b2-4ac
Δ = 12-4·1·(-399)
Δ = 1597
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1597}}{2*1}=\frac{-1-\sqrt{1597}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1597}}{2*1}=\frac{-1+\sqrt{1597}}{2}
$