Determine real number exists for relation with square roots We have $$\sqrt{x -2} = 3 -2\sqrt{x}$$.
I am to find whether a real number exists for this relation, and the real number that satisfies.
I start by squaring both sides, which yields: 
$$x - 2 = 4x - 12\sqrt{x} + 9$$.
Whence:
$$ -3x = -12\sqrt{x} + 11 \\
\sqrt{x} = \frac{x}{4} + \frac{11}{12}.
$$
But once i get here i am stuck. How can i find whether a solution exists for x from here?

Let $y=\sqrt{x}$. Then,
$$ \sqrt{x} = \frac{x}{4} + \frac{11}{12}\\
\implies 0 = \frac{y^2}{4} -y + \frac{11}{12}
$$
You should be able to use the quadratic formula to solve for $y$ (check for extraneous solutions), and then don't forget to solve for $x$ from $y$. 
