(3x-1)*(2x+4)=(3x-1)*(x-2)

Solution for (3x-1)*(2x+4)=(3x-1)*(x-2) equation:

(3x-1)(2x+4)=(3x-1)(x-2)

We move all terms to the left:

(3x-1)(2x+4)-((3x-1)(x-2))=0

We multiply parentheses ..

(+6x^2+12x-2x-4)-((3x-1)(x-2))=0


We calculate terms in parentheses: -((3x-1)(x-2)), so:

(3x-1)(x-2)

We multiply parentheses ..

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

We get rid of parentheses

3x^2-6x-1x+2

We add all the numbers together, and all the variables

3x^2-7x+2

Back to the equation:

-(3x^2-7x+2)


We get rid of parentheses

6x^2-3x^2+12x-2x+7x-4-2=0

We add all the numbers together, and all the variables

3x^2+17x-6=0

a = 3; b = 17; c = -6;
Δ = b2-4ac
Δ = 172-4·3·(-6)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-19}{2*3}=\frac{-36}{6}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+19}{2*3}=\frac{2}{6}
=1/3
$