210=n^2+n

Solution for 210=n^2+n equation:

210=n^2+n

We move all terms to the left:

210-(n^2+n)=0

We get rid of parentheses

-n^2-n+210=0

We add all the numbers together, and all the variables

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