n^2+n=1998

Solution for n^2+n=1998 equation:

n^2+n=1998

We move all terms to the left:

n^2+n-(1998)=0

a = 1; b = 1; c = -1998;
Δ = b2-4ac
Δ = 12-4·1·(-1998)
Δ = 7993
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{7993}}{2*1}=\frac{-1-\sqrt{7993}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{7993}}{2*1}=\frac{-1+\sqrt{7993}}{2}
$