(n^2+n)=240

Solution for (n^2+n)=240 equation:

(n^2+n)=240

We move all terms to the left:

(n^2+n)-(240)=0

We get rid of parentheses

n^2+n-240=0

a = 1; b = 1; c = -240;
Δ = b2-4ac
Δ = 12-4·1·(-240)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*1}=\frac{-32}{2}
=-16
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*1}=\frac{30}{2}
=15
$