How to find out whether these are the same? In the wikipedia article for Cubic equation, the root can be obtained by:

$-\frac{1}{3a}(b+C+\frac{\Delta_0}{C})$

Where $\Delta_0=b^2-3ac$ and $C=\sqrt[3]{\frac{\Delta_1\pm\sqrt{\Delta_1^2-4\Delta_0^3}}{2}}$. Also, $\Delta_1=2b^3-9abc+27a^2d$.
In another website, there's another root solution:

$$\sqrt[3]{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})+\sqrt{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}+\sqrt[3]{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})-\sqrt{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}-\frac{b}{3a}$$

I have put the latter in Wolphram|Alpha to evaluate it. the $\Delta_1$ can be seen in it; but I have no idea how to find out it and the previous solution are the same.
 A: Define:
$$\begin{align*} x_N &= -\dfrac{b}{3a} \quad \text{(average of all 3 roots, x-value of inflection point)} \\
\\
\delta^2 &= \dfrac{b^2-3ac}{9a^2} \quad \mathrm{(x \; distance^2 \; from \;} x_N \; \text{to the 2 turning points)}\\
\\
y_N &= f(x_N) = \dfrac{2b^3}{27a^2}-\dfrac{bc}{3a} +d \quad \text{(y-value of inflection point)}\\
\\
h &= 2a\delta^3 \quad \mathrm{(y \; distance \; from \;} y_N \; \text{to the 2 turning points)} \\
\end{align*}$$
(See figure 1 in this paper by Nickalls: http://www.nickalls.org/dick/papers/maths/cubic1993.pdf)
The second expression you presented can then be written as
$$x_N + \sqrt[3]{\dfrac{1}{2a}\left(-y_N + \sqrt{y_N^2 - h^2}\right) } +  \sqrt[3]{\dfrac{1}{2a}\left(-y_N - \sqrt{y_N^2 - h^2}\right) } $$
or, for $h \ne 0$,
$$x_N + \delta\left(\sqrt[3]{\dfrac{-y_N}{h} + \sqrt{\dfrac{y_N^2}{h^2} - 1 }} +  \sqrt[3]{\dfrac{-y_N}{h} - \sqrt{\dfrac{y_N^2}{h^2} - 1 }} \right) $$
In the first expression you presented, we have
$$\begin{align*} \Delta_0 & = 9a^2 \delta^2 \\
\\
\Delta_1 &= 27a^2 y_N \\
\\
C &= -3a \sqrt[3]{\dfrac{1}{2a}\left(-y_N \mp \sqrt{y_N^2 - h^2}\right) }\\
\end{align*}$$
so that expression becomes
$$ x_N + \sqrt[3]{\dfrac{1}{2a}\left(-y_N + \sqrt{y_N^2 - h^2}\right) } + \dfrac{\delta^2}{\sqrt[3]{\dfrac{1}{2a}\left(-y_N + \sqrt{y_N^2 - h^2}\right) }}$$
or, for $h \ne 0$,
$$x_N + \delta\left(\sqrt[3]{\dfrac{-y_N}{h} + \sqrt{\dfrac{y_N^2}{h^2} - 1 }} + \dfrac{1}{\sqrt[3]{\dfrac{-y_N}{h} + \sqrt{\dfrac{y_N^2}{h^2} - 1 }} }\right) $$
which after multiplying the numerator and denominator of that last term in the parentheses by $$\sqrt[3]{\dfrac{-y_N}{h} - \sqrt{\dfrac{y_N^2}{h^2} - 1 }}$$
becomes
$$x_N + \delta\left(\sqrt[3]{\dfrac{-y_N}{h} + \sqrt{\dfrac{y_N^2}{h^2} - 1 }} +  \sqrt[3]{\dfrac{-y_N}{h} - \sqrt{\dfrac{y_N^2}{h^2} - 1 }} \right) $$
So yes indeed, those two expressions for the roots of the cubic you found are equivalent.
Now I encourage you to throw away all of that classical solution for the roots of a cubic, and instead learn Nickalls' approach as presented by Nickalls and built upon by Holmes:
http://www.nickalls.org/dick/papers/maths/cubic1993.pdf
https://users.math.msu.edu/users/newhous7/math_235/lectures/cubic_gc_holmes.pdf
