Is it possible to solve the following equation for x? 
All letters (except x) are constants. 
I'm trying to create a program that can calculate the change in concentration (represented by x here) of any chemical reaction that involves 4 substances, given their initial concentrations (alpha, beta, gamma, delta) and their molar coefficient (a,b,c,d) in the balanced reaction. K represents the equilibrium constant associated with the reaction. For that, I need a formula that has x explicitly in terms of the other constants.
 A: You're basically trying to invert this equation:  given an equation for $K(x)$, you're looking for an equation for $x(K)$.  It's difficult to prove beyond the shadow of a doubt that one can't invert an equation, and usually it involves techniques that are well beyond the "algebra-precalculus" tag that you've attached to this question. However, there are a couple of techniques that can be used to get approximate results:
1. Binomial expansion approximation
If we can assume that $x$ is "small", then we can expand the expression for $K(x)$ using the binomial expansion:  $(1 + y)^n \approx 1 + ny$ for $y \ll 1$.  This means that we can approximate $K(x)$ as
\begin{align}
K &= \frac{\gamma^c \delta^d}{\alpha^a \beta^b} \left( 1 + \frac{c}{\gamma} x\right)^c \left( 1 + \frac{d}{\delta} x\right)^d \left( 1 - \frac{a}{\alpha} x\right)^{-a} \left( 1 - \frac{b}{\beta} x\right)^{-b} \\
&\approx \frac{\gamma^c \delta^d}{\alpha^a \beta^b} \left( 1 + \frac{c^2}{\gamma} x\right) \left( 1 + \frac{d^2}{\delta} x\right) \left( 1 + \frac{a^2}{\alpha} x\right) \left( 1 + \frac{b^2}{\beta} x\right) \\
&\approx \frac{\gamma^c \delta^d}{\alpha^a \beta^b} \left[ 1 + \left( \frac{a^2}{\alpha} + \frac{b^2}{\beta} + \frac{c^2}{\gamma} + \frac{d^2}{\delta} \right) x \right]
\end{align}
In this last step, we have discarded all terms proportional to $x^2$, $x^3$, etc., since if $x$ is "small", then these terms will be minuscule.  All told, then, we have
$$ x \approx \left[K \frac{\alpha^a \beta^b}{\gamma^c \delta^d} - 1 \right] \left( \frac{a^2}{\alpha} + \frac{b^2}{\beta} + \frac{c^2}{\gamma} + \frac{d^2}{\delta} \right)^{-1}.
$$
2.  Newton's method
You mention that you're "writing a program", which implies that you don't actually need an exact formula;  you just need a method that spits out a result that's accurate to a certain number of decimal places.  You're basically trying to find a root of the function
$$
f(x) = \frac{(\gamma + cx)^c (\delta + dx)^d}{(\alpha - ax)^a (\beta - bx)^b} - K,
$$ 
i.e., the value of $x$ for which $f(x) = 0$.  
Wikipedia has a thorough article on Newton's method, including pseudocode, and I encourage you to read it.  But it basically works as follows:  You start off at a particular value of $x$, denoted by $x_0$.  (In your case, $x_0 = 0$ would be a natural choice.)  You then calculate the value of $f'(x_0)$.  If $f(x_0)$ is positive, you take a "step" along the $x$-axis in the direction that the function decreases, and call that new value $x_1$;  if $f(x_0)$ is negative, you move in the direction that the function increases.  Moreover, the size of the step you take is determined by the magnitude of $f'(x_0)$;  if the function is changing rapidly, you take a small step, while if it's changing slowly, you take a big step.  
You then repeat the same procedure at $x_1$ to obtain $x_2$, and so forth until your answer is "sufficiently accurate".  Usually this means that you iterating the process until the change between $x_n$ and $x_{n+1}$ is smaller than some pre-set threshold;  you can then say that your answer is accurate to within this threshold.  For example, if you iterate your procedure until the successive answers differ by no more than $10^{-6}$, then you can be confident that your answer is within $10^{-6}$ of the "true" answer (i.e., you have 5 digits of accuracy after the decimal place.)
Other algorithms for this sort of numerical root-finding exist as well.  Newton's method is the simplest and easiest to explain, and will probably work fine for your purposes;  but if you're curious, a quick google on "numerical root-find algorithms" will give you more information than you require.
Finally, as an interesting connection:  If you start Newton's method at $x_0 = 0$, then the value of $x_1$ you'll obtain in this process is the value of $x_1$ is equal to the value of $x$ obtained from the binomial expansion.  This is because both methods involve approximating a function via the first term in its Taylor series: $f(x) \approx f(x_0) + f'(x_0) (x - x_0).$
A: [Based on the original question with all signs positive, which was edited after I posed this answer.]
Here is a construction of a nontrivial case without a solution. 
Let $a=b=c=d \ne 0$ and $\gamma = \delta$ and $\alpha = \beta$. Then you have
$$
K = (\frac{\gamma + ax}{\alpha + ax})^{2a} = (1 + \frac{\gamma -\alpha}{\alpha + ax})^{2a} 
$$ 
If $\gamma > \alpha$ and if all constants are positive, then this is decreasing with $x$  and the limit for large $x$ will be 1. So there will be no solution if $K<1$. 
Here, I presume that $x$, being the change in concentration, should be in a defining region which includes $0$. So it shouldn't be allowed to "pass" the pole and become less than $-\alpha/a$. 
