n+n^2=126

Solution for n+n^2=126 equation:

n+n^2=126

We move all terms to the left:

n+n^2-(126)=0

a = 1; b = 1; c = -126;
Δ = b2-4ac
Δ = 12-4·1·(-126)
Δ = 505
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{505}}{2*1}=\frac{-1-\sqrt{505}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{505}}{2*1}=\frac{-1+\sqrt{505}}{2}
$