n2+n=420

Solution for n2+n=420 equation:

n2+n=420

We move all terms to the left:

n2+n-(420)=0

We add all the numbers together, and all the variables

n^2+n-420=0

a = 1; b = 1; c = -420;
Δ = b2-4ac
Δ = 12-4·1·(-420)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-41}{2*1}=\frac{-42}{2}
=-21
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+41}{2*1}=\frac{40}{2}
=20
$