250=n^2+n

Solution for 250=n^2+n equation:

250=n^2+n

We move all terms to the left:

250-(n^2+n)=0

We get rid of parentheses

-n^2-n+250=0

We add all the numbers together, and all the variables

-1n^2-1n+250=0

a = -1; b = -1; c = +250;
Δ = b2-4ac
Δ = -12-4·(-1)·250
Δ = 1001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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