n(n+2)=240

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

n(n+2)=240

We move all terms to the left:

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

We multiply parentheses

n^2+2n-240=0

a = 1; b = 2; c = -240;
Δ = b2-4ac
Δ = 22-4·1·(-240)
Δ = 964
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{964}=\sqrt{4*241}=\sqrt{4}*\sqrt{241}=2\sqrt{241}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{241}}{2*1}=\frac{-2-2\sqrt{241}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{241}}{2*1}=\frac{-2+2\sqrt{241}}{2}
$