100=n(2+n)

Solution for 100=n(2+n) equation:

100=n(2+n)

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(n(n+2))+100=0


We calculate terms in parentheses: -(n(n+2)), so:

n(n+2)

We multiply parentheses

n^2+2n

Back to the equation:

-(n^2+2n)


We get rid of parentheses

-n^2-2n+100=0

We add all the numbers together, and all the variables

-1n^2-2n+100=0

a = -1; b = -2; c = +100;
Δ = b2-4ac
Δ = -22-4·(-1)·100
Δ = 404
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:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{101}}{2*-1}=\frac{2-2\sqrt{101}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{101}}{2*-1}=\frac{2+2\sqrt{101}}{-2}
$