(x-3)(x-2)=(2x-1)(x-3)

Solution for (x-3)(x-2)=(2x-1)(x-3) equation:

(x-3)(x-2)=(2x-1)(x-3)

We move all terms to the left:

(x-3)(x-2)-((2x-1)(x-3))=0

We multiply parentheses ..

(+x^2-2x-3x+6)-((2x-1)(x-3))=0


We calculate terms in parentheses: -((2x-1)(x-3)), so:

(2x-1)(x-3)

We multiply parentheses ..

(+2x^2-6x-1x+3)

We get rid of parentheses

2x^2-6x-1x+3

We add all the numbers together, and all the variables

2x^2-7x+3

Back to the equation:

-(2x^2-7x+3)


We get rid of parentheses

x^2-2x^2-2x-3x+7x+6-3=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{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*-1}=\frac{2}{-2}
=-1
$