131070=n^2+n

Solution for 131070=n^2+n equation:

131070=n^2+n

We move all terms to the left:

131070-(n^2+n)=0

We get rid of parentheses

-n^2-n+131070=0

We add all the numbers together, and all the variables

-1n^2-1n+131070=0

a = -1; b = -1; c = +131070;
Δ = b2-4ac
Δ = -12-4·(-1)·131070
Δ = 524281
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{524281}}{2*-1}=\frac{1-\sqrt{524281}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{524281}}{2*-1}=\frac{1+\sqrt{524281}}{-2}
$