Daniel,
This is one of those functions you almost have to graph in order to see what's going on. Doing so will show that there are real zeros in the intervals [2<x<1.5], [0.05<x<0], [0.5<x<1] and [2<x<2.5]. As to how to solve for these zeros, we could use the method of Newton
x_{1} = x_{0}  f(x_{0})/f'(x_{0})
where x_{0} is your "first guess" as to the real zero's value, f(x_{0}) is f(x) evaluated at that point, and f'(x_{0}) is the derivative of f(x) evaluated at x_{0}.
For example, we know from our graph that there is a real zero between 2 and  1.5. Let's push our luck and assume that that value is 1.75.
Then
x_{1} = (1.75)  {[6(x_{0})^{4}  7(x_{0})^{3}  23(x_{0})^{2} + 14(x_{0}) + 3]/[24(x_{0})^{3} 21(x_{0})^{2} 46(x_{0}) + 14]}
x_{1} = (1.75)  (1.85156)/(98.4375) = 1.73119
Now you can take the value for x1 and use it to find the next approximation, x2, by going through the process all over again.
x_{2} = x_{1}  [f(x_{1})/f'(x_{1})]
To save you some trouble, I evaluated the original expression using a TI89 Titanium calculator and here are the values I found:
x = 1.73073, or 0.169749, or .727385 or 2.33976
12/2/2013

William S.