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n(n-1)(n+1)=6
We move all terms to the left:
n(n-1)(n+1)-(6)=0
We use the square of the difference formula
n^2-1-6=0
We add all the numbers together, and all the variables
n^2-7=0
a = 1; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·1·(-7)
Δ = 28
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{7}}{2*1}=\frac{0-2\sqrt{7}}{2} =-\frac{2\sqrt{7}}{2} =-\sqrt{7} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{7}}{2*1}=\frac{0+2\sqrt{7}}{2} =\frac{2\sqrt{7}}{2} =\sqrt{7} $
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