n(n+2)=462

Solution for n(n+2)=462 equation:

n(n+2)=462

We move all terms to the left:

n(n+2)-(462)=0

We multiply parentheses

n^2+2n-462=0

a = 1; b = 2; c = -462;
Δ = b2-4ac
Δ = 22-4·1·(-462)
Δ = 1852
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:

$\sqrt{\Delta}=\sqrt{1852}=\sqrt{4*463}=\sqrt{4}*\sqrt{463}=2\sqrt{463}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{463}}{2*1}=\frac{-2-2\sqrt{463}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{463}}{2*1}=\frac{-2+2\sqrt{463}}{2}
$