2x(3x+3)=3(2x+2)

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

2x(3x+3)=3(2x+2)

We move all terms to the left:

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

We multiply parentheses

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


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

3(2x+2)

We multiply parentheses

6x+6

Back to the equation:

-(6x+6)


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
$