n(2+n)=448

Solution for n(2+n)=448 equation:

n(2+n)=448

We move all terms to the left:

n(2+n)-(448)=0

We add all the numbers together, and all the variables

n(n+2)-448=0

We multiply parentheses

n^2+2n-448=0

a = 1; b = 2; c = -448;
Δ = b2-4ac
Δ = 22-4·1·(-448)
Δ = 1796
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:

$\sqrt{\Delta}=\sqrt{1796}=\sqrt{4*449}=\sqrt{4}*\sqrt{449}=2\sqrt{449}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{449}}{2*1}=\frac{-2-2\sqrt{449}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{449}}{2*1}=\frac{-2+2\sqrt{449}}{2}
$