Are all polynomials solvable? If not, is there only a limited rangle of polynomials for which the root can be found?
Also, if $u=x^{\frac{3}{2}}+x$, is $x$ expressible in terms of $u$?
 A: Yes, there is just a limited amount of polynomials for which we can find the exact, i.e. algebraic roots by some general formula (like it's possible for quadratic polynomials). 
Particularly, the Abel-Ruffini-theorem states that there is no such general solution for polynomials of degree five or higher in terms of radicals.
This doesn't mean they don't have roots (in fact they always have), but one has to write them down or approximate them in other ways.

As to your equation (note that this is no longer a polynomial equation, which would require it's exponents to be integers, but a general power sum)
$$ u = x ^ {\frac{3}{2}} + x$$
We substitute $k = x^{\frac{1}{2}}$ and now have to solve a cubic (polynomial) equation
$$ k^3 + k^2 - u = 0 $$
For degree three, we can use a well-known but kinda unwieldy general solution formula and finally obtain zero to three results.
A: Your equation can be solved: http://www.wolframalpha.com/input/?i=solve+u%3Dx%5E%283%2F2%29%2Bx+for+x
The mathematics that decides when a polynomial is solvable by radicals is called Galois Theory and is one of the hallmarks of modern algebra.
A: The answer to your question depends entirely on what "can be found" means. I will note, before we start, that the expression you have with fractional exponents is not generally considered a polynomial: a polynomial in the variable $x$, with real coefficients, is an expression of the form
$$a_0 + a_1x + a_2x^2+ \cdots + a_nx^n$$
where $a_0,a_1,\ldots,a_n$ are real numbers, and $n$ is a nonnegative integer. Note that all the powers of the variable are integral powers. If $a_n\neq 0$, we say the polynomial has degree $n$.
When $n\leq 4$, then there are formulas that express all the roots of the polynomial in terms of the coefficients ($a_0$, $a_1,\ldots,a_n$). For $n=1$, that is a polynomial $a_0+a_1x$ with $a_1\neq 0$, the solution is simply $x=-\frac{a_0}{a_1}$. For $n=2$, that is a polynomial $a_0 + a_1x + a_2x^2$ with $a_2\neq 0$, you get the well-known quadratic formula that gives the two roots:
$$ r_1 = \frac{-a_1 + \sqrt{a_1^2 - 4a_0a_2}}{2a_2}\quad\text{and}\quad r_2 = \frac{-a_1-\sqrt{a_1^2-4a_0a_2}}{2a_2}.$$
When $n=3$ and when $n=4$ there are also formulas to express all the roots of the polynomial in terms of the coefficients and using only the operations of addition, subtractions, multiplication, division, and root extraction. For $n=3$, these are the Cardano formulae, and for $n=4$ the solution is due to Ferrari.
There are no similar formulas when $n\geq 5$; this is the celebrated Abel-Ruffini theorem. This does not mean we have no way of finding the roots, just that there is no formula that applies to all polynomials that gives the roots in terms of the coefficients, using only certain kinds of operations. For example, quintic equations (degree 5 polynomials) can be solved using other more complicated kinds of functions and operations (theta functions). 
On the other hand, there are plenty of methods for finding approximate values for roots of polynomials, or in fact values that are "as close as you want" to the roots of the polynomial. For example, Sturm's theorem can help you locate the roots approximately, and then you can combine it with Newton's method to find very good approximations to the roots of the polynomial. See for example the Wikipedia page on root-finding algorithms.  
But of course, this turns on whether "finding an approximation" qualifies as "can be found" (or even if "can be found, theoretically, given enough time to work" qualifies as "can be found"). 
For your "also" question, you want to express $x$ in terms of $u$ if $u=x^{3/2}+x$. This is slightly different from "solving a polynomial"; you are really trying to find a formula for the inverse of $x^{3/2}+x$ (assuming it has one, which it does because the function is increasing). You can turn it into a problem of solving a polynomial because if you let $z=x^{1/2}$, then you have $u=z^3+z^2$, which is equivalent to $z^3+z^2-u=0$; this is a cubic polynomial, so the roots of the polynomial can be expressed in terms of the coefficients (in this case, $1$, $1$, $0$, and $-u$) using Cardano's formulae. Then you would replace $x^{1/2}$ for $z$ in the case in which $z$ is nonnegative and real, and squaring gives you the answer. So, yes, it can be done.
