44=n(3n-1)

Solution for 44=n(3n-1) equation:

44=n(3n-1)

We move all terms to the left:

44-(n(3n-1))=0


We calculate terms in parentheses: -(n(3n-1)), so:

n(3n-1)

We multiply parentheses

3n^2-1n

Back to the equation:

-(3n^2-1n)


We get rid of parentheses

-3n^2+1n+44=0

We add all the numbers together, and all the variables

-3n^2+n+44=0

a = -3; b = 1; c = +44;
Δ = b2-4ac
Δ = 12-4·(-3)·44
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-23}{2*-3}=\frac{-24}{-6}
=+4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+23}{2*-3}=\frac{22}{-6}
=-3+2/3
$