(2x-1)(3x-5)=(4x-2)(3x-2)

Solution for (2x-1)(3x-5)=(4x-2)(3x-2) equation:

(2x-1)(3x-5)=(4x-2)(3x-2)

We move all terms to the left:

(2x-1)(3x-5)-((4x-2)(3x-2))=0

We multiply parentheses ..

(+6x^2-10x-3x+5)-((4x-2)(3x-2))=0


We calculate terms in parentheses: -((4x-2)(3x-2)), so:

(4x-2)(3x-2)

We multiply parentheses ..

(+12x^2-8x-6x+4)

We get rid of parentheses

12x^2-8x-6x+4

We add all the numbers together, and all the variables

12x^2-14x+4

Back to the equation:

-(12x^2-14x+4)


We get rid of parentheses

6x^2-12x^2-10x-3x+14x+5-4=0

We add all the numbers together, and all the variables

-6x^2+x+1=0

a = -6; b = 1; c = +1;
Δ = b2-4ac
Δ = 12-4·(-6)·1
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*-6}=\frac{-6}{-12}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*-6}=\frac{4}{-12}
=-1/3
$