Turning points of particle 
A particle of mass $m$ and energy $E<0$ moves in a one-dimensional Morse potential: 
$$V(x)=V_0(e^{-2ax}-2e^{-ax}),\qquad       V_0,a>0,\qquad E>-V_0.$$ 

Determine the turning points of the movement and the period of the oscillation of the particle. 

I have started learning for my exam and this is one of the exercises in my textbook. 
Never dealt with this type of question, so these are my thoughts so far: 
To get the turning points I was thinking of solving the equation $E=V_0(e^{-2ax}-2e^{-ax})$. I was doing the arithmetics with the absolute value of $E$. But still I couldn't seem to find the values for $x$. At the end I used Wolfram Alpha to find the values but it gave me results with complex values. Is there a simple way to solve this type of equations for $x$? 
Anyway, about the period, I assume it's the time it takes for the particle to get from $x_1$ to $x_2$. But how am I supposed to approach this? How would I get a time value just out of the equation for the potential? 
I hope someone can help me out here. 
 A: Energy conservation dictates
$$ E = \frac{1}{2}m\dot{x}^2 + V(x) = \text{const}$$
With some arithmetic it follows
$$ \dot{x} = \frac{dx}{dt} = \sqrt{2m^{-1}(E-V(x))}$$
This ODE can be solved via separation of variables,
yielding
$$ \int_{t_1}^{t_2}dt = \int_{x_1}^{x_2} \frac{dx}{\sqrt{2m^{-1}(E-V(x))}}$$
The integral on the left hand side can be evaluated immediately, where $t_1$ and $t_2$ are understood as the times when the particle is at $x_1$ or $x_2$ respectively. So it is simply half the period. 
Observe that the integral on the right diverges when $x$ approaches the turning point $E=V(x)$. 
This method of solving Newtons equations in a 1d potential should be treated in any textbook on mechanics.
For a general potential it is in general hard or impossible to find the turning points in closed form. Here however, you can substitute $y=\exp(-ax)$ and solve the corresponding quadratic equation. I'll leave the explicit calculation to you.
A: nephente's answer solves for the period. My answer is just made to make you see how to go on from the point you were, solving:
$$ e^{-2ax}-2e^{-ax}=\frac{E}{V_0} $$
We make the change $y=e^{-ax}$, which yields:
$$y^2-2y-\frac{E}{V_0} = 0$$
Second grade equation. Solve for $y$, and have in mind that $E/V_0$ is a negative number. It may give you complex solutions otherwise.
