(2x-7)(3-2x)=(3-2x)(5-3x)

Solution for (2x-7)(3-2x)=(3-2x)(5-3x) equation:

(2x-7)(3-2x)=(3-2x)(5-3x)

We move all terms to the left:

(2x-7)(3-2x)-((3-2x)(5-3x))=0

We add all the numbers together, and all the variables

(2x-7)(-2x+3)-((-2x+3)(-3x+5))=0

We multiply parentheses ..

(-4x^2+6x+14x-21)-((-2x+3)(-3x+5))=0


We calculate terms in parentheses: -((-2x+3)(-3x+5)), so:

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

We multiply parentheses ..

(+6x^2-10x-9x+15)

We get rid of parentheses

6x^2-10x-9x+15

We add all the numbers together, and all the variables

6x^2-19x+15

Back to the equation:

-(6x^2-19x+15)


We get rid of parentheses

-4x^2-6x^2+6x+14x+19x-21-15=0

We add all the numbers together, and all the variables

-10x^2+39x-36=0

a = -10; b = 39; c = -36;
Δ = b2-4ac
Δ = 392-4·(-10)·(-36)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-9}{2*-10}=\frac{-48}{-20}
=2+2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+9}{2*-10}=\frac{-30}{-20}
=1+1/2
$