<s>[INST] I can't find the other two solutions to this equation. As an exercise, I have to use Cardano's formula
$$
x^3 = px + q$$
$$x = \sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} - \frac{p^3}{27}}} + \sqrt[3]{\frac{q}{2}-\sqrt{\frac{q^2}{4} - \frac{p^3}{27}}}
$$
to solve the equation 
$
x^3 = 15x+4.
$ 
I finally get
$$ x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}$$
$$ x = 4 $$
but I have no idea how to find the other two solutions to the equation, and the procedures I've found googling are not employing Cardano's formula.
Thank you for your help.
 [/INST] Since the degree is 3, if you have a solution, you can make a factorisation and have a second degree equation to solve : 
$$\begin{align*}
x^3&=15x+4 \tag{1}\\
4^3&=15\times 4 +4\tag{2}
\end{align*}
$$
with (1)-(2) :
$$x^3-4^3=15x-15\times 4 $$
so (with $a^3-b^3=(a-b)(a^2+ab+b^2)$ )
$$ (x-4)(x^2 + 4x +16)= 15(x-4) \quad  \Leftrightarrow \quad (x-4)(x^2 + 4x +16-15)= 0$$
and you just have to solve then 
$$ x^2 +4x +1$$
to find the others solutions :
$$ x=-2\pm \sqrt{3}.$$ 
</s>