Finding the Roots of this Cubic I need to find the roots of the cubic $x^3+19x^2-109x-2,431=0$ in order to continue on finding the roots of a Quartic.
But here, I am stuck. I don't really know any good ways to find the roots of any cubic with integer solutions.
Trying Cardano's Method gave me something completely different than what Desmos is telling me. The roots are supposedly $11,-13, -17$. So I'm wondering if it's possible to find the solutions to any cubic without knowledge on the value of the roots.
 A: As Ross Millikan answered, you probably made a mistake in your calculations using Cardano's method.
Let us do it exactly as the method is described here
$$\Delta = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$$ Using $a=1$, $b=19$, $c=-109$, $d=-2431$, this gives $\Delta =7225344$ and so, three real roots.
$$\Delta_0 = b^2 - 3ac=688$$
$$\Delta_1 = 2b^3 -  9abc + 27a^2d=-33280$$
$$\Delta_1^2-4 \Delta_0^3= -195084288$$
$$\sqrt{\Delta_1^2-4 \Delta_0^3}=8064 i \sqrt{3}$$
$$C = \sqrt[3]{\frac{\Delta_1 \pm \sqrt{{\Delta_1}^2 - 4 {\Delta_0}^3}}2}=4 \left(4\pm3 i \sqrt{3}\right)$$
One of the roots is given by
$$x = - \frac{1}{3a}\left(b+C+\frac{\Delta_0}{C}\right)$$ Using $C=4 \left(4+3 i \sqrt{3}\right)$, this gives $$x=-\frac{1}{3} \left(19+\frac{172}{4+3 i \sqrt{3}}+4 \left(4+3 i
   \sqrt{3}\right)\right)=-17$$
Factoring let us with
$$x^3+19x^2-109x-2431=(x+17)(x^2+2 x-143)=0$$ and the quadratic is easy to solve for the other roots $(x=11,x=-13)$.
A: As you've observed, the roots are 11, −13, and −17 and they can be found in several ways. Cardano's method works as well but might produce results that appear complicated. Using my own implementation of Cardano's method in Mathematica, I find the roots to be:
\begin{align}
x_1 &=-\frac{19}{3}+\sqrt[3]{\frac{16640}{27}-\frac{448
   i}{\sqrt{3}}}+\sqrt[3]{\frac{16640}{27}+\frac{448
   i}{\sqrt{3}}} \\
x_2 &= -\frac{19}{3}+\sqrt[3]{\frac{16640}{27}+\frac{448
   i}{\sqrt{3}}} \left(-\frac{1}{2}-\frac{i
   \sqrt{3}}{2}\right)+\sqrt[3]{\frac{16640}{27}-\frac{448
   i}{\sqrt{3}}} \left(-\frac{1}{2}+\frac{i
   \sqrt{3}}{2}\right) \\
x_3 &= -\frac{19}{3}+\sqrt[3]{\frac{16640}{27}-\frac{448
   i}{\sqrt{3}}} \left(-\frac{1}{2}-\frac{i
   \sqrt{3}}{2}\right)+\sqrt[3]{\frac{16640}{27}+\frac{448
   i}{\sqrt{3}}} \left(-\frac{1}{2}+\frac{i
   \sqrt{3}}{2}\right)
\end{align}
While it might not look like it at first glance, these happen to be 11, −13, and −17 in disguise.
A: The roots you quote are correct per Alpha. Cardano's method will find the roots without prior knowledge, so you must have made a mistake. This will also yield to the rational root theorem.
