(n^2+n)/2=5050

Solution for (n^2+n)/2=5050 equation:

(n^2+n)/2=5050

We move all terms to the left:

(n^2+n)/2-(5050)=0

We multiply all the terms by the denominator


(n^2+n)-5050*2=0

We add all the numbers together, and all the variables

(n^2+n)-10100=0

We get rid of parentheses

n^2+n-10100=0

a = 1; b = 1; c = -10100;
Δ = b2-4ac
Δ = 12-4·1·(-10100)
Δ = 40401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{40401}=201$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-201}{2*1}=\frac{-202}{2}
=-101
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+201}{2*1}=\frac{200}{2}
=100
$