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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+15x^2-12x-25x+20)-((3x-2)(3x-2))=0


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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

9x^2-12x+4

Back to the equation:

-(9x^2-12x+4)


We get rid of parentheses

15x^2-9x^2-12x-25x+12x+20-4=0

We add all the numbers together, and all the variables

6x^2-25x+16=0

a = 6; b = -25; c = +16;
Δ = b2-4ac
Δ = -252-4·6·16
Δ = 241
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{-(-25)-\sqrt{241}}{2*6}=\frac{25-\sqrt{241}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{241}}{2*6}=\frac{25+\sqrt{241}}{12}
$