(x-5)(3x-3)=3(x-9)+6x

Solution for (x-5)(3x-3)=3(x-9)+6x equation:

(x-5)(3x-3)=3(x-9)+6x

We move all terms to the left:

(x-5)(3x-3)-(3(x-9)+6x)=0

We multiply parentheses ..

(+3x^2-3x-15x+15)-(3(x-9)+6x)=0


We calculate terms in parentheses: -(3(x-9)+6x), so:

3(x-9)+6x

We add all the numbers together, and all the variables

6x+3(x-9)

We multiply parentheses

6x+3x-27

We add all the numbers together, and all the variables

9x-27

Back to the equation:

-(9x-27)


We get rid of parentheses

3x^2-3x-15x-9x+15+27=0

We add all the numbers together, and all the variables

3x^2-27x+42=0

a = 3; b = -27; c = +42;
Δ = b2-4ac
Δ = -272-4·3·42
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-15}{2*3}=\frac{12}{6}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+15}{2*3}=\frac{42}{6}
=7
$