n(n+2)=116

Solution for n(n+2)=116 equation:

n(n+2)=116

We move all terms to the left:

n(n+2)-(116)=0

We multiply parentheses

n^2+2n-116=0

a = 1; b = 2; c = -116;
Δ = b2-4ac
Δ = 22-4·1·(-116)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{13}}{2*1}=\frac{-2-6\sqrt{13}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{13}}{2*1}=\frac{-2+6\sqrt{13}}{2}
$