Solve the equation ${\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}} = 9x-3$ I tried factoring the expression inside the square root, but that does not seem to help. Squaring the equation makes it even more terrible.

Can anyone provide a hint about what should be done?
Notice that

$$\left({\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}}\right) \left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) = 9x-3.$$
Thus, 

$$9x-3 =  \left(9x-3\right)\left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) \implies $$

$$9x-3 = 0 \implies x=1/3$$ or

$${\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}} = 1.$$

But the minimal value of $x^2-x+1$ is $3/4$, which implies that 
$$ 2{\sqrt {x^2-x+1}} \geq \sqrt{3} &gt; 1.$$

Therefore $x =1/3$ is the only real solution.
Move $2\sqrt {x^2 - x + 1}$ to the rhs, then square both sides.