100=(n+1)(n)

Solution for 100=(n+1)(n) equation:

100=(n+1)(n)

We move all terms to the left:

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


We calculate terms in parentheses: -((n+1)n), so:

(n+1)n

We multiply parentheses

n^2+n

Back to the equation:

-(n^2+n)


We get rid of parentheses

-n^2-n+100=0

We add all the numbers together, and all the variables

-1n^2-1n+100=0

a = -1; b = -1; c = +100;
Δ = b2-4ac
Δ = -12-4·(-1)·100
Δ = 401
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{401}}{2*-1}=\frac{1-\sqrt{401}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{401}}{2*-1}=\frac{1+\sqrt{401}}{-2}
$