(n^2)+n=90

Solution for (n^2)+n=90 equation:

(n^2)+n=90

We move all terms to the left:

(n^2)+n-(90)=0

a = 1; b = 1; c = -90;
Δ = b2-4ac
Δ = 12-4·1·(-90)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*1}=\frac{-20}{2}
=-10
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*1}=\frac{18}{2}
=9
$