20=n(n-3)

Solution for 20=n(n-3) equation:

20=n(n-3)

We move all terms to the left:

20-(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+20=0

We add all the numbers together, and all the variables

-1n^2+3n+20=0

a = -1; b = 3; c = +20;
Δ = b2-4ac
Δ = 32-4·(-1)·20
Δ = 89
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{-(3)-\sqrt{89}}{2*-1}=\frac{-3-\sqrt{89}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{89}}{2*-1}=\frac{-3+\sqrt{89}}{-2}
$