Htykuut

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

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{-bpmsqrt{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{-bpmsqrt{b^2-4ac}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{(-6)^2-4(3)(-6)}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{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?

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:
$$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$

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

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

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