45=n(2n-1)

Solution for 45=n(2n-1) equation:

45=n(2n-1)

We move all terms to the left:

45-(n(2n-1))=0


We calculate terms in parentheses: -(n(2n-1)), so:

n(2n-1)

We multiply parentheses

2n^2-1n

Back to the equation:

-(2n^2-1n)


We get rid of parentheses

-2n^2+1n+45=0

We add all the numbers together, and all the variables

-2n^2+n+45=0

a = -2; b = 1; c = +45;
Δ = b2-4ac
Δ = 12-4·(-2)·45
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*-2}=\frac{-20}{-4}
=+5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*-2}=\frac{18}{-4}
=-4+1/2
$