n(n+1)(n-1)=1320

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Solution for n(n+1)(n-1)=1320 equation:



n(n+1)(n-1)=1320
We move all terms to the left:
n(n+1)(n-1)-(1320)=0
We use the square of the difference formula
n^2-1-1320=0
We add all the numbers together, and all the variables
n^2-1321=0
a = 1; b = 0; c = -1321;
Δ = b2-4ac
Δ = 02-4·1·(-1321)
Δ = 5284
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:
$\sqrt{\Delta}=\sqrt{5284}=\sqrt{4*1321}=\sqrt{4}*\sqrt{1321}=2\sqrt{1321}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1321}}{2*1}=\frac{0-2\sqrt{1321}}{2} =-\frac{2\sqrt{1321}}{2} =-\sqrt{1321} $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1321}}{2*1}=\frac{0+2\sqrt{1321}}{2} =\frac{2\sqrt{1321}}{2} =\sqrt{1321} $

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