Q factor for an LRC series circuit
The resonant frequency is the geometric mean of the upper and lower half power frequencies, not the arithmetic mean you seem to be using here.

\$\omega_0 = \sqrt{\omega_l \omega_h} \ne \dfrac{\omega_l + \omega_h}{2}\$

That's probably all you need but here's some additional information just in case.

[UPDATE:  upon further reflection, there's something else you need.  The relationship between Q and BW you are using is a high-Q approximation.  Deriving the genuine Q from this approximation requires a bit more than I've given below.]

For a series RLC, we have:

\$ Z = \dfrac{1 + j \omega RC - \omega^2LC}{j \omega C}\$

The numerator is of the form:

\$ 1 + j\frac{1}{Q} \frac{\omega}{\omega_0} - (\frac{\omega}{\omega_0})^2\$

Comparing these two forms, see that:

\$ \omega_0 = \frac{1}{\sqrt{LC}}\$

\$ Q = \dfrac{\sqrt{\frac{L}{C}}}{R}\$

How to see that the resonant frequency is the geometric mean?

Consider the product of two first order functions:

\$ (1 + j\dfrac{\omega}{\omega_l})(1 + j\dfrac{\omega}{\omega_h}) = 1 + j \omega (\dfrac{1}{\omega_l} + \dfrac{1}{\omega_h}) - \dfrac{\omega^2}{\omega_l \omega_h}\$

From this, we deduce:

\$\omega_0 = \sqrt{\omega_l \omega_h}\$

\$ Q \omega_0 = \omega_l || \omega_h =\dfrac{1}{\dfrac{1}{\omega_l} + \dfrac{1}{\omega_h}} \rightarrow Q = \dfrac{\omega_0}{\omega_l + \omega_h}\$