n(n+1)=133590

Solution for n(n+1)=133590 equation:

n(n+1)=133590

We move all terms to the left:

n(n+1)-(133590)=0

We multiply parentheses

n^2+n-133590=0

a = 1; b = 1; c = -133590;
Δ = b2-4ac
Δ = 12-4·1·(-133590)
Δ = 534361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{534361}=731$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-731}{2*1}=\frac{-732}{2}
=-366
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+731}{2*1}=\frac{730}{2}
=365
$