<s>[INST] How to solve for a non-factorable cubic equation? I want to know how one would go about solving an unfactorable cubic. I know how to factor cubics to solve them, but I do not know what to do if I cannot factor it. For example, if I have to solve for $x$ in the cubic equation:
$$2x^3+6x^2-x+4=0$$
how would I do it?
Edit: I have heard people telling me to convert it into a depressed cubic (where the $x^2$ term disappears), but I have no idea how to do that.
Edit 2: I am aware that there is a cubic formula, that for any cubic equation $ax^3+bx^2+cx+d$, it's roots are:
$$x = \sqrt[3]{-\dfrac{b^3}{27a^3} + \dfrac{bc}{6a^2} - \dfrac{d}{2a} + \sqrt{\left(-\dfrac{b^3}{27a^3} + \dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 - \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} + \sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a} - \sqrt{\left(-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 - \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} - \dfrac{b}{3a}$$
This formula is way too complicated so I do not even bother memorizing it or using it.
 [/INST] Observe that $2x^3+6x^2 = 2(x^3+3x^2) = 2(x+1)^3 - 6x-2$. Now let $X=x+1$, and divising by 3, your equation is now something like $X^3+pX+t=0$, which can be solved by Cardano method.
</s>