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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+15x^2-12x-5x+4)-((3x-5)(3x-1))=0


We calculate terms in parentheses: -((3x-5)(3x-1)), so:

(3x-5)(3x-1)

We multiply parentheses ..

(+9x^2-3x-15x+5)

We get rid of parentheses

9x^2-3x-15x+5

We add all the numbers together, and all the variables

9x^2-18x+5

Back to the equation:

-(9x^2-18x+5)


We get rid of parentheses

15x^2-9x^2-12x-5x+18x+4-5=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
$