Please run the following code

    from sympy.solvers import solvefrom sympy import Symbolx = Symbol('x')R2 = solve(-109*x**5/3870720+4157*x**4/1935360-3607*x**3/69120+23069*x**2/60480+5491*x/2520+38-67,x)print R2

The output of the code is

[2*CRootOf(109*x**5 - 4157*x**4 + 50498*x**3 - 184552*x**2 -527136*x + 3507840, 0), 2*CRootOf(109*x**5 - 4157*x**4 + 50498*x**3- 184552*x**2 - 527136*x + 3507840, 1), 2*CRootOf(109*x**5 - 4157*x**4 + 50498*x**3 - 184552*x**2 - 527136*x + 3507840, 2),2*CRootOf(109*x**5 - 4157*x**4 + 50498*x**3 - 184552*x**2 -527136*x + 3507840, 3), 2*CRootOf(109*x**5 - 4157*x**4 + 50498*x**3- 184552*x**2 - 527136*x + 3507840, 4)]

Can someone explain what the answer represent and how to get the output in conventional form i.e. say if the answer is 0.1,0.2,0.3,0.1,0.4 sympy usually outputs the answer as [0.1,0.2,0.3,0.1,0.4]

To get numerical approximations in an answer, you can use N(). Since you have multiple solutions, you can loop through the list. I have used an easier equation since yours takes a while ...

Try this:

from sympy.solvers import solvefrom sympy import Symbol, N
x = Symbol('x')
#R2 = solve(-109*x**5/3870720+4157*x**4/1935360-3607*x**3/69120+23069*x**2/60480+5491*x/2520+38-67,x)
R2 = solve(x**2+2*x-4,x)
print R2
print [N(solution) for solution in R2]

[EDIT]: As mentioned in the comments below, the fifth order equation can only be solved after upgrading sympy (to 1.0 in my case).