3x(x-9)=(2x-3)(x-9)

Solution for 3x(x-9)=(2x-3)(x-9) equation:

3x(x-9)=(2x-3)(x-9)

We move all terms to the left:

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

We multiply parentheses

3x^2-27x-((2x-3)(x-9))=0

We multiply parentheses ..

3x^2-((+2x^2-18x-3x+27))-27x=0


We calculate terms in parentheses: -((+2x^2-18x-3x+27)), so:

(+2x^2-18x-3x+27)

We get rid of parentheses

2x^2-18x-3x+27

We add all the numbers together, and all the variables

2x^2-21x+27

Back to the equation:

-(2x^2-21x+27)


We add all the numbers together, and all the variables

3x^2-27x-(2x^2-21x+27)=0

We get rid of parentheses

3x^2-2x^2-27x+21x-27=0

We add all the numbers together, and all the variables

x^2-6x-27=0

a = 1; b = -6; c = -27;
Δ = b2-4ac
Δ = -62-4·1·(-27)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12}{2*1}=\frac{18}{2}
=9
$