Wolfram is right, and it's doable. I'll show you how:

Let's divide by x^2 (we'll lose 2 imaginary roots this way, but we don't need those anyway).

Let's rearrange the equation a bit, so we get the following:

Now let's say that
Then y^2 would be:
That looks quite similar to the first part of our equation, we just need to subtract 2 and we'll have what we need, which looks like this:

Now, we have a plain quadratic. Solve it using the quadratic formula, then substitute back to get the two roots of x we were looking for (2+-square root 3). Check it yourself if you don't believe me :P Event is over.