19998=n(n+1)

Solution for 19998=n(n+1) equation:

19998=n(n+1)

We move all terms to the left:

19998-(n(n+1))=0


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

n(n+1)

We multiply parentheses

n^2+n

Back to the equation:

-(n^2+n)


We get rid of parentheses

-n^2-n+19998=0

We add all the numbers together, and all the variables

-1n^2-1n+19998=0

a = -1; b = -1; c = +19998;
Δ = b2-4ac
Δ = -12-4·(-1)·19998
Δ = 79993
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{79993}}{2*-1}=\frac{1-\sqrt{79993}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{79993}}{2*-1}=\frac{1+\sqrt{79993}}{-2}
$