How to solve for a non-factorable cubic equation?

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.

To depress a cubic means to write it in the form $y^3+py+q=0$ by performing a convenient substitution. It is not hard, and I'll give you a hint on how to do it yourself.

Hint. Show that every cubic equation of the form $x^3+ax^2+bx+c=0$ can be written as $y^3+py+q=0$ by performing a substitution $x=y-w$. Expand the binomials and find $w$ by letting the coefficient of $y^2$ be zero.

Once you depress a cubic, you have to solve the simpler equation $y^3+py+q=0$. The solution is given by Cardano's formula, a way simpler than the one you stated: $$y=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}.$$ In fact, if you write the equation as $z^3+3zm=2n$, the formula is even simpler: $$z=\sqrt[3]{n+\sqrt{n^2+m^3}}+\sqrt[3]{n-\sqrt{n^2+m^3}}.$$ It is not easy to come up with this, but you can see a somewhat simple proof here. This equation motivated the first appearance of complex numbers and has a very curious history, so make sure you read it!

There is a very acessible historical reading in the first chapter of Needham's book on complex analysis. Highly recommended.