Algebra

The only restriction for x is that the value inside the square root must be greater than 0. 

 

We can write

\(6-x-x^2-2x^2 \geq 0\)

 

Solving for x, we have

\(-3x^2-x+6 \geq 0\)

 

Completing the square, we have

\(-3\left(x+\frac{1}{6}\right)^2+\frac{73}{12}\ge \:0\)

\(-3\left(x+\frac{1}{6}\right)^2\ge \:-\frac{73}{12}\)

\(\left(x+\frac{1}{6}\right)^2\le \frac{73}{36}\)

 

Square rooting both sides,  we get

\(-\sqrt{\frac{73}{36}}\le \:x+\frac{1}{6}\le \sqrt{\frac{73}{36}}\)

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

 

So the domain is \(\frac{-\sqrt{73}-1}{6}\le \:x\le \frac{\sqrt{73}-1}{6}\)

 

Thanks! :)