(6n-5)=n(3n-2)

Solution for (6n-5)=n(3n-2) equation:

(6n-5)=n(3n-2)

We move all terms to the left:

(6n-5)-(n(3n-2))=0

We get rid of parentheses

6n-(n(3n-2))-5=0


We calculate terms in parentheses: -(n(3n-2)), so:

n(3n-2)

We multiply parentheses

3n^2-2n

Back to the equation:

-(3n^2-2n)


We get rid of parentheses

-3n^2+6n+2n-5=0

We add all the numbers together, and all the variables

-3n^2+8n-5=0

a = -3; b = 8; c = -5;
Δ = b2-4ac
Δ = 82-4·(-3)·(-5)
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{-(8)-2}{2*-3}=\frac{-10}{-6}
=1+2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*-3}=\frac{-6}{-6}
=1
$