20=n(n-1)

Solution for 20=n(n-1) equation:

20=n(n-1)

We move all terms to the left:

20-(n(n-1))=0


We calculate terms in parentheses: -(n(n-1)), so:

n(n-1)

We multiply parentheses

n^2-1n

Back to the equation:

-(n^2-1n)


We get rid of parentheses

-n^2+1n+20=0

We add all the numbers together, and all the variables

-1n^2+n+20=0

a = -1; b = 1; c = +20;
Δ = b2-4ac
Δ = 12-4·(-1)·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*-1}=\frac{-10}{-2}
=+5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*-1}=\frac{8}{-2}
=-4
$