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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

3-2(2x+2)

determiningTheFunctionDomain
-2(2x+2)+3

We multiply parentheses

-4x-4+3

We add all the numbers together, and all the variables

-4x-1

Back to the equation:

-(-4x-1)


We get rid of parentheses

6x^2+2x+4x+1=0

We add all the numbers together, and all the variables

6x^2+6x+1=0

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

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3}}{2*6}=\frac{-6-2\sqrt{3}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3}}{2*6}=\frac{-6+2\sqrt{3}}{12}
$