Algebra

The only limit for the problem we have is that

$6-x-x^2-2x^2\geq0$ so there are no imaginary numbers,

Combining like terms, we have

$-3x^2-x+6\geq0$

Applying the quadratic equation, we have

$\frac{-\sqrt{73}-1}{6}\le \:x\le \frac{\sqrt{73}-1}{6}$

So our answer is 

$\frac{-\sqrt{73}-1}{6}\le \:x\le \frac{\sqrt{73}-1}{6}$

Not the clearest explenation, I can elaborate if needed. 

Thanks! :)