Difference between real and complex solutions of cubic equations Take for an example, this equation.
$$x^3+15x+4=0$$
This equation has two complex solutions and a real one.
$$x≈0.1327-3.8798i$$
$$x≈0.1327+3.8798i$$
$$x≈-0.26542$$
What's extra in the complex solution compared to the real solution? 
Does the two kinds of solutions have any difference in physical significance? 
 A: There is no physical significance as long as you don't relate the equation to some physical phenomenon.
For instance, if the polynomial was the characteristic polynomial of a linear differential equation, complex roots indicate an oscillatory behavior rather than pure damping.
"Benefits" of complex solutions is that they always exist, whatever the polynomial; their use is even mandated to compute the real roots of some cubic polynomials. "Disadvantages" are that you need to switch to complex algebra.
But you have no choice, complex roots are not negotiable.

Every polynomial with real coefficients can be factored as a product of quadratic trinomials and a single linear binomial (in case of odd degree), all with real coefficients. The linear binomial always gives a real root.
For the quadratic trinomials, you can rewrite them by completing the square, i.e.
$$x^2+\alpha x+\beta=(x-a)^2+b$$ which corresponds to a parabola. The apex of the parabola has the abscissa $x=a$, and the ordinate $y=b$. When $b$ is negative (blue), the parabola crosses the axis and yields two roots. When $b$ is positive (green), it yields two imaginary roots
$$x-a=\pm i\sqrt{b},$$
and the absolute value of $b$ tells you how "far" you are from real roots.

A: Your equation factors to
$$x^3 - 15x - 4 = 0 \Leftrightarrow (x-4)(x^2 +4x + 1) = 0 $$
This means there are roots where $x-4 = 0$ or $x^2 + 4x + 1 = 0$.
(And as people have pointed out, there are no imaginary roots here.)
There are no general objective "advantages"/"disadvantages" to having real or complex roots. The only thing I can think of is that if you plot the graph, you can by visual inspection see the number of real roots and their approximate value.
A: Not sure, whether this hits your intention :
The complex solutions have a theoretical advantage. We can be sure that a polynomial with degree $n$ (non-real coefficients allowed) has exactly $n$ roots (where roots of multiplicity $k$ are counted $k$ times).
In particular, every non-constant complex polynomial has a complex root. This is not the case in the reals. Such a field is called "algebraically closed".
A: It has $3$ real solutions http://www.wolframalpha.com/input/?i=x%5E3-15x-4%3D0
To find the roots: This is equivalent to $x^3-15x-4=0$. Use rational root theorem and try: $\pm 1, \pm 2, \pm 4$ to get that $4$ is a root. Then use polynomial long division to get a quadratic that can be solved by quadratic formula. 
A: in some sense , it may be thought of vector of numbers , it gives number a direction , but that's obviously a disputable claim. but some properties to be aware of are that if polynomials have real coefficients complex roots always come as conjugates . for real part , sum of real part will be negative of the ratio of coefficients of  $x^{n-1}$ and $x^n$ where $n$ is the maximum power of $x$. both of these can be checked in your solution.   
