Using substitution in complex equation $x^{10}-2x^5+\frac{4}{3}=0$
I substituted
$t=x^5$
Then I got with the quadratic formula the result
$t_1=1+\frac{\sqrt\frac{4}{3}}{2}i$
$t_2=1-\frac{\sqrt\frac{4}{3}}{2}i$  
How do I calculate the roots now?
When I now resubstitute with  
$t=x^5$  
and form the polar coordinates do I get the roots for the whole equation?
 A: You now need to solve $$x^5 = \left(1 \pm \frac{i}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}}\exp\left({i\left(\frac{\pi}{6} + 2\pi k\right)}\right).$$
This should net you ten solutions, $5$ from each - which is intuitively right since you were originally solving a polynomial of degree $10$.
So $$x = \left(\frac{2}{\sqrt{3}}\right)^{1/5}\exp{\left(i\left(\frac{\pi}{30} + \frac{2\pi k}{5}\right)\right)}$$
Then taking various values of $k$ will get you all your roots. 
A: With the quadratic formula, you should have gotten $t = 1 \pm \dfrac{1}{\sqrt{3}}i$, which is half of what you got.
After this, you need to solve $x^5 = 1 + \dfrac{1}{\sqrt{3}}i$ and $x^5 = 1 - \dfrac{1}{\sqrt{3}}i$. Each of these will give you $5$ roots, for a total of $10$ roots. This makes sense since you were finding the roots of a polynomial of degree $10$.
A: You are correct, you will get the roots.
However, note that when substituting $x^5$ back in and taking the $5$-th root, that you will get $5$ different answers.
Let's start with $t_1$.$$t_1=1+\frac{\sqrt\frac{4}{3}}{2}i$$$$(x_1)^5=1+\frac{\sqrt\frac{4}{3}}{2}i$$$$1+\frac{\sqrt\frac{4}{3}}{2}i=1+\sqrt{\frac13}i=2\sqrt{\frac13}e^{i\theta}$$$$\theta_1\approx0.523598776$$$$\theta_2=\theta_1+2\pi$$$$\theta_3=\theta_1+4\pi$$$$\theta_4=\theta_1+6\pi$$$$\theta_5=\theta_1+8\pi$$$$(x_1)^5=2\sqrt{\frac13}e^{i\theta}$$$$x_1=\sqrt[10]{\frac43}e^{\frac{i\theta}5}$$Substitute your $\theta$ values in for your roots.
Finding $x_2$ is done similarly.  Cutting to it, it simply has the same of everything except $\theta_{t_2}=-\theta_{t_1}$
Which allows us to have all ten roots in one small equation:$$x=\sqrt[10]{\frac43}e^{\pm\frac{i\theta}5}$$
