n^2+n=1980

Solution for n^2+n=1980 equation:

n^2+n=1980

We move all terms to the left:

n^2+n-(1980)=0

a = 1; b = 1; c = -1980;
Δ = b2-4ac
Δ = 12-4·1·(-1980)
Δ = 7921
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7921}=89$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-89}{2*1}=\frac{-90}{2}
=-45
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+89}{2*1}=\frac{88}{2}
=44
$