Yeah, that works, though you either omitted a step or got lucky when solving your integral for x. Notice that your integral is negative everywhere--plugging ~1.386 in you get -1/2, not 1/2.

We know P(X > x0) = Integral of f from x0 to infinity, basically by definition of the probability density. This integral is -e^(-x/2) *evaluated between x0 and infinity*. Then P(X > x0) = [-e^(-inf/2) + e^(-x0/2)] = e^(-x0/2)=1/2. You had -e^(-x0/2)=1/2. The exact solution happens to be 2ln2.