1000=n^2+2n

Solution for 1000=n^2+2n equation:

1000=n^2+2n

We move all terms to the left:

1000-(n^2+2n)=0

We get rid of parentheses

-n^2-2n+1000=0

We add all the numbers together, and all the variables

-1n^2-2n+1000=0

a = -1; b = -2; c = +1000;
Δ = b2-4ac
Δ = -22-4·(-1)·1000
Δ = 4004
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:

$\sqrt{\Delta}=\sqrt{4004}=\sqrt{4*1001}=\sqrt{4}*\sqrt{1001}=2\sqrt{1001}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{1001}}{2*-1}=\frac{2-2\sqrt{1001}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{1001}}{2*-1}=\frac{2+2\sqrt{1001}}{-2}
$