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

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

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

We move all terms to the left:

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

We multiply parentheses

3x^2-9x-((2x-4)(x-3))=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((+2x^2-6x-4x+12)), so:

(+2x^2-6x-4x+12)

We get rid of parentheses

2x^2-6x-4x+12

We add all the numbers together, and all the variables

2x^2-10x+12

Back to the equation:

-(2x^2-10x+12)


We add all the numbers together, and all the variables

3x^2-9x-(2x^2-10x+12)=0

We get rid of parentheses

3x^2-2x^2-9x+10x-12=0

We add all the numbers together, and all the variables

x^2+x-12=0

a = 1; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·1·(-12)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*1}=\frac{6}{2}
=3
$