# Cubic polynomial with 1 real root and 2 complex conjugated roots (real coefficients)

I am stuck on this problem about cubic polynomials. I rely on the Wikipedia page on the topic. Using wikipedia notations (chapter "General formula for roots") :

For the case where $\Delta > 0$, there are 3 real roots and I could check that I can find them.

But for the case where $\Delta < 0$, when there is one real root and 2 complex conjugated roots, I can not find the real root using the formula they give : $$x_k = -\frac{1}{3a}\left( b + u_kC+\bar{u_k}\bar{C}\right)$$ where : $$C = \sqrt[3]{\frac{\Delta_1 + \sqrt{-27a^2\Delta}}{2}}$$ and $u_k$ are the three complex cube roots of unity.

Actually, when $\Delta <0$, $C$ is a real number, and to me the sum $u_kC + \bar{u_k}\bar{C} = 2C\Re(u_k)$ is always real. So where are the complex congugated roots ? And also, depending on the $u_k$ chosen, there is a factor -2 in the root founded.

Can somebody help me to find where I am going wrong ?

I got it ! It was a notation issue.

In Cardano's formula, $\bar{C}$, in the case of one real root and two complex conjugated roots, is not the conjugated of $C$ but another real value :

$$C = \sqrt[3]{\frac{\Delta_1 + \sqrt{-27a^2\Delta}}{2}}$$

and

$$\bar{C} = \sqrt[3]{\frac{\Delta_1 - \sqrt{-27a^2\Delta}}{2}}$$

And the formula gives all the roots of the polynomial using the three cube roots of unity.

On a general case, depending on the sign of $\Delta$, $C$ and $\bar{C}$ can be two real or two complex conjugated numbers, which gives a coherency to the formula.