0=n2+n-1300

Solution for 0=n2+n-1300 equation:

0=n2+n-1300

We move all terms to the left:

0-(n2+n-1300)=0

We add all the numbers together, and all the variables

-(+n^2+n-1300)+0=0

We add all the numbers together, and all the variables

-(+n^2+n-1300)=0

We get rid of parentheses

-n^2-n+1300=0

We add all the numbers together, and all the variables

-1n^2-1n+1300=0

a = -1; b = -1; c = +1300;
Δ = b2-4ac
Δ = -12-4·(-1)·1300
Δ = 5201
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{5201}}{2*-1}=\frac{1-\sqrt{5201}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{5201}}{2*-1}=\frac{1+\sqrt{5201}}{-2}
$