n(n-1)=36

Solution for n(n-1)=36 equation:

n(n-1)=36

We move all terms to the left:

n(n-1)-(36)=0

We multiply parentheses

n^2-1n-36=0

a = 1; b = -1; c = -36;
Δ = b2-4ac
Δ = -12-4·1·(-36)
Δ = 145
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{145}}{2*1}=\frac{1-\sqrt{145}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{145}}{2*1}=\frac{1+\sqrt{145}}{2}
$