Polynomial equation, cannot solve for $x$

Question

$$3x^2-4x-4+x^3=x^3+2x+2$$

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?

Answer 1

so we start off with: $$x^3+3x^2-4x-4=x^3+2x+2$$ which we can re-write as: $$3x^2-6x-6=0$$ so we know that: $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{6\pm\sqrt{36-(4)(3)(-6)}}{6}=1\pm\frac{\sqrt{108}}{6}=1\pm\sqrt{3}$$

The mistake you made was $36-72$ instead of $36+72$

Answer 2

$$x^3+3x^2-4x-4=x^3+2x+2$$ $$3x^2-6x-6=0$$ $$x^2-2x-2=0$$

$$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-2)}}{2(1)}$$ $$x=\frac{2\pm\sqrt{4+8}}{2}=\frac{2\pm 2\sqrt 3}{2}=1\pm\sqrt 3$$