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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(6x(x+1)), so:

6x(x+1)

We multiply parentheses

6x^2+6x

Back to the equation:

-(6x^2+6x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-6x^2+6=0

a = -6; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-6)·6
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{-(0)-12}{2*-6}=\frac{-12}{-12}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-6}=\frac{12}{-12}
=-1
$