(n)=n*(n-1)

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

(n)=n(n-1)

We move all terms to the left:

(n)-(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+n+1n=0

We add all the numbers together, and all the variables

-1n^2+2n=0

a = -1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-1}=\frac{-4}{-2}
=+2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-1}=\frac{0}{-2}
=0
$