I am not an expert of complex numbers, and in a (non-English) wiki article I found what seems a very simple way of solving a cubic when $\Delta ( {\tfrac {q^{2}}{4}}+{\tfrac {p^{3}}{27}}) <0: $
($
y=\sqrt[3]{-\frac{q}{2}+i\sqrt{-\Delta}}+\sqrt[3]{-\frac{q}{2}- i\sqrt{-\Delta}} $) :
$$1) y_1=u_1+v_1=2\sqrt{-\frac{p}{3}}\cdot\cos\frac{\theta}{3} $$ $$ 2) y_{2}=u_{2}+v_{3}=2{\sqrt {-{\frac {p}{3}}}}\cdot \cos {\frac {\theta +2\pi }{3}} $$ $$3) y_{3}=u_{3}+v_{2}=2{\sqrt {-{\frac {p}{3}}}}\cdot \cos {\frac {\theta +4\pi }{3}} $$
If someone can provide a simple way for determining the angle $\theta$, considering for example the cubic: $$x^3-15x-4=0$$ it would be almost magical to multiply $2\sqrt{5}$ by its ($1/3$,… etc) cosine and find the three solutions in a nanosecond.
Can you explain or refer me to a link where such procedure is explained?
Can anyone tell howthe angle $\theta$ is representedand derived on the complex plane from $p,q$?
Using the information from here, it's easy. So, assuming that $\Delta<0$, the roots of the equation $x^3+px+q=0$ are$$2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}p}\right)-\frac{2\pi k}3\right),$$with $k\in\{0,1,2\}$. In your specific case, $\frac{3q}{2p}\sqrt{\frac{-3}p}=\frac2{5\sqrt5}$ and therefore the roots are$$2\sqrt5\cos\left(\frac13\arccos\left(\frac2{5\sqrt5}\right)-\frac{2\pi k}3\right),$$again with $k\in\{0,1,2\}$. This is equal to $4$, $-2+\sqrt3$, and $-2-\sqrt3$ respectively.