Cubic Formula Doesn't Seem to Work for $x^3+3x^2-50x-52=0$ So I'll jump right into it. I've derived the following formula set for solving cubics of the form $ax^3+bx^2+cx+d=0$:
$$z_{k\pm}=\sqrt[3]{\frac{9abc-2b^3-27a^2d}{54a^3}\pm\sqrt{\frac{4ac^3+27a^2d^2-18abcd-b^2c^2+4b^3d}{108a^4}}}$$
$$x=z_{k\pm}-\frac{\left(\frac{-b^2}{3a^2}+\frac{c}{a}\right)}{3z_{k\pm}}-\frac{b}{3a}$$
The idea is that you solve for the 6 $z$-values defined by $k=0, 1, 2$ and each of these $k$-values has a positive and negative result based upon the selection for the $\pm$ sign. Also note that this derivation is heavily based upon this article (http://math.sfsu.edu/smith/Documents/Cubic&Quartic.pdf). The article dictates that these 6 $z$-values translate to only 3 distinct $x$-values (since cubics only have 3 solutions).
This all went smoothly and I was able to successfully show the viability of the formula set by solving the cubic $-2x^3+3x^2-x+5=0$. The formula was successful in capturing all 3 $x$-values without any issues.
However, upon further inspection of the formula, I have noticed an issue when the radicand of the square root is negative. This would make the radicand of the cube root imaginary, thus leaving $z$ with no real solutions; hence, $x$ would have no real solutions when $\frac{4ac^3+27a^2d^2-18abcd-b^2c^2+b^3d}{108a^4}\lt0$. Of course this cannot be true since every cubic has at least one real solution.
To test this theory I tried to solve
$$x^3+3x^2-50x-52=0$$
using the formula. This cubic does indeed make the radicand of the square root negative and happens to have three real roots: $-1, -1-\sqrt{52}, -1+\sqrt{52}$. However, upon solving the formula for these values, $z_{0+}$ and $z_{0-}$ were complex and $z_{1+}$ gave a domain error on my TI-84+ calculator. My reference article provides some insight into dealing with a negative radicand, but it doesn't make sense to me. Any help is greatly appreciated!
 A: For negative radicands, the 6 $z$-values translates into 3 distinctive $x$-values as well.
Take the example $x^3+3x^2-50x-52=0$ given in the post, $z$‘s are evaluated as
$$z_{k\pm} =\sqrt[3]{\pm \sqrt{-\frac{53^3}{27}}}= \pm\sqrt{\frac{53}3}e^{i \frac{1+4\pi k}6},\>\>\>k=0,1,2$$
and the corresponding $x$’s are
$$x_{k\pm} = z_{k\pm} +\frac{53}{3z_{k\pm}}-1
= z_{k\pm} +{z_{k\mp}}-1=2Re(z_{k\pm})-1
$$
Explicitly, the 6 $x$-values are
\begin{align}
x_{0+} & = 2\sqrt{\frac{53}3}\cos\frac\pi6-1=\sqrt{53}-1\\
x_{0-} & = -2\sqrt{\frac{53}3}\cos\frac\pi6-1=-\sqrt{53}-1\\
x_{1+} & = 2\sqrt{\frac{53}3}\cos\frac{5\pi}6-1=-\sqrt{53}-1\\
x_{1-} & = -2\sqrt{\frac{53}3}\cos\frac{5\pi}6-1=\sqrt{53}-1\\
x_{2+} & = 2\sqrt{\frac{53}3}\cos\frac{3\pi}2-1=-1\\
 x_{2-} & =- 2\sqrt{\frac{53}3}\cos\frac{3\pi}2-1=-1\\
\end{align}
of which, only 3 values, $-1$, $\sqrt{53}-1$ and $-\sqrt{53}-1$, are distinctive, as expected.
A: This is called "casus irreducibilis". To solve it, you need to use a different function. $1$ way to do this is with trigonometry. $$ax^{3}+bx^{2}+cx+d=0,\ x=\frac{2\sqrt{b^{2}-3ac}\cos\left(\frac{1}{3}\cos^{-1}\left(-\frac{b^{3}-4.5abc+13.5a^{2}d}{\sqrt{\left(b^{2}-3ac\right)^{3}}}\right)\right)-b}{3a}$$ This is basically the expression that Quanto generates by hand in his derivation. If you want to get radical expressions, use the rational root theorem to get $1$ solution, then solve for the other $2$. $$x_{2,\ 3}=\frac{\sqrt{b^{2}-4ac-3a^{2}x_{1}^{2}-2abx_{1}}-b-ax_{1}}{2a}$$ Side note: The form you got the formula in is more complicated than it needs to be. An equivalent simplified form is $$ax^{3}+bx^{2}+cx+d=0,\ x=\frac{\sqrt[3]{\sqrt{\left(b^{3}-4.5abc+13.5a^{2}d\right)^{2}-\left(b^{2}-3ac\right)^{3}}-\left(b^{3}-4.5abc+13.5a^{2}d\right)}-\sqrt[3]{\sqrt{\left(b^{3}-4.5abc+13.5a^{2}d\right)^{2}-\left(b^{2}-3ac\right)^{3}}+\left(b^{3}-4.5abc+13.5a^{2}d\right)}-b}{3a}$$
