<s>[INST] Polynomial equation, cannot solve for $x$ 
$$3x^2-4x-4+x^3=x^3+2x+2$$

This boils down to (I think):
$$3x^2 - 6x - 6 = 0$$
I'm trying to solve for $x$ using the polynomial equation:
$$\begin{align}
x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\
a &= 3 \\[0.2ex]
b &= -6 \\
c &= -6
\end{align}$$
My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}$. I'm unable to replicate this.
Here's my work:
$$\begin{align}
x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\[0.8ex]
x&=\frac{-6\pm\sqrt{(-6)^2-4(3)(-6)}}{2a} \\[0.8ex]
x&=\frac{-6\pm\sqrt{36-72}}{2a}
\end{align}$$
Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.
Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into 
$$3x^2-6x-6=0$$
How can I arrive at solutions for $x$ where $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}\,$ per my textbook's given solutions?
 [/INST] $$x^3+3x^2-4x-4=x^3+2x+2$$
$$3x^2-6x-6=0$$
$$x^2-2x-2=0$$
$$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
$$x=\frac{2\pm\sqrt{4+8}}{2}=\frac{2\pm 2\sqrt 3}{2}=1\pm\sqrt 3$$
</s>