0=n^2+n-1000

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Solution for 0=n^2+n-1000 equation:



0=n^2+n-1000
We move all terms to the left:
0-(n^2+n-1000)=0
We add all the numbers together, and all the variables
-(n^2+n-1000)=0
We get rid of parentheses
-n^2-n+1000=0
We add all the numbers together, and all the variables
-1n^2-1n+1000=0
a = -1; b = -1; c = +1000;
Δ = b2-4ac
Δ = -12-4·(-1)·1000
Δ = 4001
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{4001}}{2*-1}=\frac{1-\sqrt{4001}}{-2} $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{4001}}{2*-1}=\frac{1+\sqrt{4001}}{-2} $

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