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.

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 :
$$ x=-2\pm \sqrt{3}.$$ 
