(3x-2)=(3x-2)(2x-3)

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

(3x-2)=(3x-2)(2x-3)

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

-((+6x^2-9x-4x+6))+3x-2=0


We calculate terms in parentheses: -((+6x^2-9x-4x+6)), so:

(+6x^2-9x-4x+6)

We get rid of parentheses

6x^2-9x-4x+6

We add all the numbers together, and all the variables

6x^2-13x+6

Back to the equation:

-(6x^2-13x+6)


We add all the numbers together, and all the variables

3x-(6x^2-13x+6)-2=0

We get rid of parentheses

-6x^2+3x+13x-6-2=0

We add all the numbers together, and all the variables

-6x^2+16x-8=0

a = -6; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·(-6)·(-8)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*-6}=\frac{-24}{-12}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*-6}=\frac{-8}{-12}
=2/3
$