According to Wikipedia, a depressed cubic has only one root and 2 imaginary roots. Is this true? Can a depressed cubic of the form $x^3+px+q=0$ have 2 or 3 real roots?
Edit: Here is a screenshot of the Wikipedia excerpt: 
[Math] Depressed cubic roots
cubicspolynomialsroots
Best Answer
Yes, it can. Consider the polynomial having as roots $-3$, $1$ and $2$: since the sum of the roots is zero, the coefficient of $x^2$ is zero (Viète's formula): $$ (x+3)(x-1)(x-2)=0. $$
One can also have two roots (one is repeated, of course): $$ (x-4)(x+2)^2=0 $$
Or just one, if two roots are complex: $$ (x-2)(x+1-i)(x+1+i)=0. $$
Check doing the multiplications, if you don't trust Viète.
However, I can't find what you say in the linked Wikipedia page.