I'm trying to use Cardano's method to solve this equation: $$x^3+6x=20 \tag{1}$$
As described on Wikipedia, I let $x = u + v$ and expand in $(1)$: $$(u+v)^3+6(u+v)=20$$ $$u^3 + v^3 + (3uv+6)(u+v)-20=0 \tag{2}$$
I then let $3uv + 6 = 0$ and substitute in $(2)$: $$u^3 + v^3 - 20 = 0$$ $$u^3 + v^3 = 20 \tag{3}$$
I also express $uv$ as a product of cubics: $$3uv + 6 = 0$$ $$uv = -2$$ $$u^3v^3 = -8$$ $$-4u^3v^3 = 32 \tag{4}$$
At this point, Wikipedia says "the combination of these two equations [$(3)$ and $(4)$] leads to a quadratic equation" which I think I can also be achieved by squaring $(3)$ and adding $(4)$ to both sides: $$u^6 + 2u^3v^3 + v^6 = 400$$ $$u^6 - 2u^3v^3 + v^6 = 432$$ $$(u^3 - v^3)^2 = 432$$ $$u^3 - v^3 = \pm 12\sqrt{3} \tag{5}$$
I then get $u$ by adding $(3)$ and $(5)$: $$2u^3 = 20 + 12\sqrt{3} \textrm{ or } 20 - 12\sqrt{3}$$ $$u = \sqrt[3]{10 + 6\sqrt{3}} \textrm{ or } \sqrt[3]{10 - 6\sqrt{3}}$$
and $v$ by subtracting $(3)$ and $(5)$: $$2v^3 = 20 - 12\sqrt{3} \textrm{ or } 20 + 12\sqrt{3}$$ $$v = \sqrt[3]{10 - 6\sqrt{3}} \textrm{ or } \sqrt[3]{10 + 6\sqrt{3}}$$
I finally get $x$ by adding $u$ and $v$: $$x = \sqrt[3]{10 + 6\sqrt{3}} + \sqrt[3]{10 - 6\sqrt{3}}$$ $$\textrm{ or } \sqrt[3]{10 - 6\sqrt{3}} + \sqrt[3]{10 + 6\sqrt{3}}$$ $$\textrm{ or } 2\sqrt[3]{10 + 6\sqrt{3}}$$ $$\textrm{ or } 2\sqrt[3]{10 - 6\sqrt{3}}$$
I know there's a real solution, so that only leaves $x = 2\sqrt[3]{10 + 6\sqrt{3}}$ which equals approximately $195$ instead of $20$ in the original equation. I can only find the correct real solution by using $x = u - v$ instead of $x = u + v$:
$$x = \sqrt[3]{10 + 6\sqrt{3}} - \sqrt[3]{6\sqrt{3} - 10}$$
So, am I misusing Cardano's method somehow?
You forgot the condition $u^3v^3=-8$. The correct solution is the first you enunciated.
That said, your way of solving the system of equations $\; \begin{cases}u^3+v^3=20\\u^3v^3=-8\end{cases}\;$ is over complicated.
Just use what any high-school student knows to solve the problem of finding two numbers the sum $s$ and product $p$ of which are given: they're roots of the quadratic equation (if any): $$t^2-st+p=t^2-20t-8=0$$ The reduced discriminant is $\Delta'=100+8$, hence the roots $$u^3,v^3=10\pm\sqrt{108}=10\pm6\sqrt 3.$$ You must take both roots, because the problem involves the condition on the product is $8$ – i.e. it implies the qudratic equation, but is not equivalent to it. Thus $$x=u+v=\sqrt[3]{10-6\sqrt 3}+\sqrt[3]{10+6\sqrt 3}.$$
Edit:
As pointed out by @André Nicolas $10\pm6\sqrt 3=(1\pm\sqrt3)^3$, so that the root is equal to $2$.