-999(n^2+1)=2000n

Solution for -999(n^2+1)=2000n equation:

-999(n^2+1)=2000n

We move all terms to the left:

-999(n^2+1)-(2000n)=0

We add all the numbers together, and all the variables

-2000n-999(n^2+1)=0

We multiply parentheses

-999n^2-2000n-999=0

a = -999; b = -2000; c = -999;
Δ = b2-4ac
Δ = -20002-4·(-999)·(-999)
Δ = 7996
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:

$\sqrt{\Delta}=\sqrt{7996}=\sqrt{4*1999}=\sqrt{4}*\sqrt{1999}=2\sqrt{1999}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2000)-2\sqrt{1999}}{2*-999}=\frac{2000-2\sqrt{1999}}{-1998}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2000)+2\sqrt{1999}}{2*-999}=\frac{2000+2\sqrt{1999}}{-1998}
$