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! :)