How to solve the equation $x^2+2=4\sqrt{x^3+1}$? From the Leningrad Mathematical Olympiad, 1975:

Solve $x^2+2=4\sqrt{x^3+1}$.

In answer sheet only written $x=4+2\sqrt{3}\pm \sqrt{34+20\sqrt{3}}$. 
How to solve this?

I think the "fastest" way to deal with this is surely to square both members, to get
$$x^4 -16x^3 +4x^2 - 12 = 0$$
Now you can treat is as the quartic equation it is, via Ferrari's method for quartic equations.
I will write you down the general procedure for a quartic of the form
$$ax^4 + bx^3 + cx^2 + dx + e = 0$$
And leave you the easy maths behind.
Define the following quantities
$$P = 8ac - 3b^2$$
$$Q = b^3 + 8da - 4abc$$
$$R = c^2 - 3bd + 12 ae$$
$$S = 64a^3e - 16a^2c^2 + 16ab^2c - 16 a^2bd - 3b^4$$
Define other elements
$$U = \frac{P}{8a^2}$$
$$V = \frac{Q}{8a^3}$$
$$Y = 2c^3 - 9bcd + 27b^2e - 27ad^2 - 27ace$$
$$Z = \sqrt[3]{\frac{Y + \sqrt{Y^2 - 4R^2}}{2}}$$
$$W = \frac{1}{2} \sqrt{-\frac{2U}{3} + \frac{1}{3a} \left( Z + \frac{R}{Z}\right) }$$
Final Solutions (note: only two of them will be real, in your case)
$$x_{1,\ 2} = -\frac{b}{4a} - W \pm \sqrt{-4W^2 - 2U + \frac{V}{W}}$$
$$x_{3,\ 4} = -\frac{b}{4a} + W \pm \sqrt{-4W^2 - 2U - \frac{V}{W}}$$
NOTICE In your case $\color{red}{d = 0}$ which simplifies a bit the resolvent.
