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

Solution for 3(x-2)(x+1)=(3x-6)(3x+3) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(3x-6)(3x+3)

We multiply parentheses ..

(+9x^2+9x-18x-18)

We get rid of parentheses

9x^2+9x-18x-18

We add all the numbers together, and all the variables

9x^2-9x-18

Back to the equation:

-(9x^2-9x-18)


We multiply parentheses

3x^2+3x-6x-(9x^2-9x-18)-6=0

We get rid of parentheses

3x^2-9x^2+3x-6x+9x+18-6=0

We add all the numbers together, and all the variables

-6x^2+6x+12=0

a = -6; b = 6; c = +12;
Δ = b2-4ac
Δ = 62-4·(-6)·12
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*-6}=\frac{-24}{-12}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*-6}=\frac{12}{-12}
=-1
$