This question is the same (or similar) as this one but in that question, the answer "use the trigonometric method" is not explained further in the Wikipedia page (it was 6 years ago so the page has probably been heavily edited since this question) so it needs to be asked again or at least edited up to date.
Take, for example, equation $x^3 – 7x + 5 = 0$
This is the graph of the cubic curve:
The equation clearly has three real roots. However, the equation is not factorable, so to solve it we would have to resort to a cubic formula. Let's use the mainstream cubic formula (not depressed) here: Eventually after substituting $a = 1$, $b = 0$, $c = -7$ and $d = 5$ into all three solutions we get (from WolframAlpha):
Even though all three solutions are real solutions, they all have complex closed forms (in their non-trigonometric states), a separate question I asked a few hours ago explained how to convert some specific closed forms into their trigonometric forms using Euler's formula (which I now understand) – but is this 'rule' applicable to all examples of real numbers with seemingly complex closed expressions? With enough time, could I theoretically remove the imaginary numbers from all three real solutions, and in their place substitute matching trigonometric identities?
Best Answer
The answer is Yes.
$$x^3 - 7x + 5 = 0$$
Now, follow the steps given here for $a=1$, $b=0$, $c=-7$ and $d=5$.
You have $$\Delta=697 \qquad \qquad p=7\qquad \qquad q=5$$
$\Delta >0$ means three real roots. So, apply the trigonometric solution to obtain $$x_k=2 \sqrt{\frac{7}{3}} \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(-\frac{15 }{14}\sqrt{\frac{3}{7}}\right)\right)\right)\quad \text{for} \quad k=0,1,2$$
Isn't a bit nicer than what Wolfram Alpha gave ?
Edit
Let me do the same for $$x^3 + 7x + 5 = 0$$
Follow the steps for $a=1$, $b=0$, $c=7$ and $d=5$.
You have $$\Delta=-2047 \qquad \qquad p=7\qquad \qquad q=5$$
$\Delta <0$ means one real roots. So, apply the hyperbolic solution to obtain $$x_r=-2 \sqrt{\frac{7}{3}} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{15}{14} \sqrt{\frac{3}{7}}\right)\right)$$