3x^2+3(x+4)x=2

Solution for 3x^2+3(x+4)x=2 equation:

3x^2+3(x+4)x=2

We move all terms to the left:

3x^2+3(x+4)x-(2)=0

We multiply parentheses

3x^2+3x^2+12x-2=0

We add all the numbers together, and all the variables

6x^2+12x-2=0

a = 6; b = 12; c = -2;
Δ = b2-4ac
Δ = 122-4·6·(-2)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*6}=\frac{-12-8\sqrt{3}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*6}=\frac{-12+8\sqrt{3}}{12}
$