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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

(x-3)(-3x+3)

We multiply parentheses ..

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

We get rid of parentheses

-3x^2+3x+9x-9

We add all the numbers together, and all the variables

-3x^2+12x-9

Back to the equation:

-(-3x^2+12x-9)


We get rid of parentheses

-4x^2+3x^2+6x+4x-12x-6+9=0

We add all the numbers together, and all the variables

-1x^2-2x+3=0

a = -1; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-1)·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*-1}=\frac{-2}{-2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*-1}=\frac{6}{-2}
=-3
$