n2+n=210

Solution for n2+n=210 equation:

n2+n=210

We move all terms to the left:

n2+n-(210)=0

We add all the numbers together, and all the variables

n^2+n-210=0

a = 1; b = 1; c = -210;
Δ = b2-4ac
Δ = 12-4·1·(-210)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-29}{2*1}=\frac{-30}{2}
=-15
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+29}{2*1}=\frac{28}{2}
=14
$