$2x^3-2x^2-3x+2=0$ has 3 real root, but they are all express in such way:
$x=\dfrac{1}{3}\left(1+\dfrac{\sqrt[3]{-23+3i\sqrt{237}}}{\sqrt[3]{2^2}}+\dfrac{11}{\sqrt[3]{2(-23+3i\sqrt{237}})}\right)$
it is in complex format.
there is a way like this:
$x=\dfrac{1}{3}+\dfrac{\sqrt{22}}{3}\cos{\dfrac{\arccos{\dfrac{-23}{11\sqrt{22}}}}{3}}$
it is in real format.
but if I want to explain to some one that the root like this can't be constructed by straight and compass,I had problem.
if the root $x$ is shown in cubic root ,such as $\sqrt[3]{2+\sqrt{23}}$, then it is easy to say. but for a angle of $\dfrac{\alpha}{3}$, there is some possibility to construct depends on $\alpha$.
I try to convert $\dfrac{\sqrt{22}}{3}\cos{\dfrac{\arccos{\dfrac{-23}{11\sqrt{22}}}}{3}}$ into $\sqrt[3]{p+\sqrt{q}}$ but I fall into a loop of cubic root so I can't find such a way.
Is it means that there is no possibility to find such expression ?
then how can I explain this root can't be constructed by straight and compass?