2x(3x-1)=3(x+1)

Solution for 2x(3x-1)=3(x+1) equation:

2x(3x-1)=3(x+1)

We move all terms to the left:

2x(3x-1)-(3(x+1))=0

We multiply parentheses

6x^2-2x-(3(x+1))=0


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

3(x+1)

We multiply parentheses

3x+3

Back to the equation:

-(3x+3)


We get rid of parentheses

6x^2-2x-3x-3=0

We add all the numbers together, and all the variables

6x^2-5x-3=0

a = 6; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·6·(-3)
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{97}}{2*6}=\frac{5-\sqrt{97}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{97}}{2*6}=\frac{5+\sqrt{97}}{12}
$