n(n+1)(n-1)=210

Solution for n(n+1)(n-1)=210 equation:

n(n+1)(n-1)=210

We move all terms to the left:

n(n+1)(n-1)-(210)=0

We use the square of the difference formula

n^2-1-210=0

We add all the numbers together, and all the variables

n^2-211=0

a = 1; b = 0; c = -211;
Δ = b2-4ac
Δ = 02-4·1·(-211)
Δ = 844
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{844}=\sqrt{4*211}=\sqrt{4}*\sqrt{211}=2\sqrt{211}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{211}}{2*1}=\frac{0-2\sqrt{211}}{2}
=-\frac{2\sqrt{211}}{2}
=-\sqrt{211}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{211}}{2*1}=\frac{0+2\sqrt{211}}{2}
=\frac{2\sqrt{211}}{2}
=\sqrt{211}
$