How to find the root of this polynomial equation of third degree? x^3 -3x + 4 = 0 

How do I find roots of the above equation?
I have tried plugging in values of x but it is not satisfying the equation above.
 A: The full solution to the cubic is:
$$x = (q + u)^{\frac{1}{3}}   +   (q - u)^{\frac{1}{3}}   +   p$$
$$u = \sqrt{q^2 + (r-p^2)^3}$$
$$p = -\frac{b}{3a}$$
$$q = p^3 + \frac{bc-3ad}{6a^2}$$
$$r = \frac{c}{3a}$$
Here we have $a=1$, $b=0$, $c=-3$, and $d=4$. Hence:
$$r = \frac{-3}{3} = -1$$
$$p = -\frac{0}{3} = 0$$
$$q = 0^3 + \frac{0-3(4)}{6} = \frac{-12}{6} = -2$$
$$u = \sqrt{(-2)^2 + (-1)^3} = \sqrt{3}$$
$$x = (\sqrt{3}-2)^{\frac{1}{3}}   +   (\sqrt{3}-2)^{\frac{1}{3}}$$
$$\therefore x= 2 (\sqrt{3}-2)^{\frac{1}{3}}$$
There are also two complex solutions, but to get these simply divide your original polynomial by the solution we get, yielding a quadratic. An application of the quadratic formula will yield the other two solutions.
A: There is no "nice" solution.
You can show that there are no rational solutions by using the Rational Root Theorem, the only possible rational roots are $\pm 1, \pm 2, \pm 4$, none of which satisfy the equation.
The only solution methods left are to use the full cubic solution (Cardano's method), which is messy, but gives an exact solution in radicals; and a numerical method, which is quick and easy, but does not give an exact solution.
In practice, numerical methods are applied most commonly in real life situations where cubics and quartics crop up.
If you just want the solution without worrying about the details, use an online solver like this one: http://www.1728.org/cubic.htm
Your equation has one negative real root and a pair of conjugate complex roots.
