n(n-1)=8

Solution for n(n-1)=8 equation:

n(n-1)=8

We move all terms to the left:

n(n-1)-(8)=0

We multiply parentheses

n^2-1n-8=0

a = 1; b = -1; c = -8;
Δ = b2-4ac
Δ = -12-4·1·(-8)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{33}}{2*1}=\frac{1-\sqrt{33}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{33}}{2*1}=\frac{1+\sqrt{33}}{2}
$