108=n(n-3)

Solution for 108=n(n-3) equation:

108=n(n-3)

We move all terms to the left:

108-(n(n-3))=0


We calculate terms in parentheses: -(n(n-3)), so:

n(n-3)

We multiply parentheses

n^2-3n

Back to the equation:

-(n^2-3n)


We get rid of parentheses

-n^2+3n+108=0

We add all the numbers together, and all the variables

-1n^2+3n+108=0

a = -1; b = 3; c = +108;
Δ = b2-4ac
Δ = 32-4·(-1)·108
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*-1}=\frac{-24}{-2}
=+12
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*-1}=\frac{18}{-2}
=-9
$