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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We multiply parentheses ..

4x^2-((+10x^2-4x-15x+6))-6x=0


We calculate terms in parentheses: -((+10x^2-4x-15x+6)), so:

(+10x^2-4x-15x+6)

We get rid of parentheses

10x^2-4x-15x+6

We add all the numbers together, and all the variables

10x^2-19x+6

Back to the equation:

-(10x^2-19x+6)


We add all the numbers together, and all the variables

4x^2-6x-(10x^2-19x+6)=0

We get rid of parentheses

4x^2-10x^2-6x+19x-6=0

We add all the numbers together, and all the variables

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

a = -6; b = 13; c = -6;
Δ = b2-4ac
Δ = 132-4·(-6)·(-6)
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{-(13)-5}{2*-6}=\frac{-18}{-12}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*-6}=\frac{-8}{-12}
=2/3
$