x^2+3x+3(-4x+x^2+x)=6

Solution for x^2+3x+3(-4x+x^2+x)=6 equation:

x^2+3x+3(-4x+x^2+x)=6

We move all terms to the left:

x^2+3x+3(-4x+x^2+x)-(6)=0

We multiply parentheses

x^2+3x^2-12x+3x+3x-6=0

We add all the numbers together, and all the variables

4x^2-6x-6=0

a = 4; b = -6; c = -6;
Δ = b2-4ac
Δ = -62-4·4·(-6)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{33}}{2*4}=\frac{6-2\sqrt{33}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{33}}{2*4}=\frac{6+2\sqrt{33}}{8}
$