Roots of a polynomial

How many solutions does the following equation have with positive real part?
z^4 + 8z^3 + 3z^2 + 8z + 3 = 0

I figured someone might have some fun with this problem since it requires so little knowledge to at least understand. I currently know of two solutions, but neither is obvious/elementary.

Here's a tiny bit of discussion to get things started. z here is a complex number; obviously if z is real and positive the left side is positive so there are no solutions in that case, so any potential solutions with positive real part must have non-zero imaginary part. From the fundamental theorem of calculus, there are 4 solutions (counted with multiplicity) overall. Since the coefficients are all real numbers, the solutions with non-zero imaginary part come in complex conjugate pairs. Thus there are 0, 2, or 4 solutions with positive real part. If you graph the function for real numbers, you find that it has precisely two solutions (both of course negative)--this fact can be justified rigorously without computation using some basic calculus and the intermediate value theorem. So, there are 0 or 2 solutions with positive real part. Which is it?

*The answer is (spoilerdontreadifyoudontwanttoknowtheansweristwothereisaidit).

Hmm, I'll write it down and do it at school since I have got nothing better to do there. Expect a reply from me in +- 6 Hours.

I wanted to mention I did find an elementary proof, though it's a bit tedious. Its only remarkable feature is that in both it and my much shorter complex analytic proof, the final step is showing that x^2 - 3x + 3 > 0 for all real x. I'm not sure if there's a connection or if this is just a strange little coincidence. The two quadratics arise from very different manipulations so I lean towards coincidence. I haven't investigated further.