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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(2(3x+9)), so:

2(3x+9)

We multiply parentheses

6x+18

Back to the equation:

-(6x+18)


We get rid of parentheses

6x^2-6x-6x-18=0

We add all the numbers together, and all the variables

6x^2-12x-18=0

a = 6; b = -12; c = -18;
Δ = b2-4ac
Δ = -122-4·6·(-18)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-24}{2*6}=\frac{-12}{12}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+24}{2*6}=\frac{36}{12}
=3
$