Finding the frequencies for a given phase difference between supply voltage and supply current
I see no problem solving the two quadratic formulas.

You are seeking the half-power points.

$$f_1, f_2 = f_0 \left ( \sqrt {1 + \frac {1}{4Q^2}}\ \mp\ \frac {1}{2Q}\right  ) \ \ \ \ [1]$$
Q = 2 and \$f_0\$ = 1591.549Hz.

So \$f_1 = 1591.549Hz \times \left ( \sqrt {1 + \frac {1}{4 \times 2^2}}\ -\ \frac {1}{2 \times 2}\right )\$  = 1242.644Hz and \$f_2\$ = 2038.419Hz.

Or:
$$f_1, f_2 = f_0 \left ( \sqrt {\left ({\frac {R}{2L}} \right )^2 + \frac {1}{LC}}\ \mp\ \frac {R}{2L}\right ) \ \ \ \ [2]$$

Resonance Chapter

The rest is not part of the answer, but lots of references do not show how the formulas are derived.  

Deriving half-power points from quality factor.

Quality factor:
$$Q = \frac {f_0}{BW} = \frac {f_0}{f_2\ -\ f_1} = \frac {1}{R} \sqrt {\frac {L}{C}} = \frac {\omega_0 \ L}{R}$$

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

At \$\theta = 45^{\circ}, Z = \sqrt {2} R\$.

$$\sqrt {R^2 +\left ({\omega L - \frac {1}{\omega C}} \right )^2} = \sqrt {2} R$$
$$R^2 +\left ({\omega L - \frac {1}{\omega C}} \right )^2 = 2 R^2$$
$$\left ({\omega L - \frac {1}{\omega C}}\right )^2 = R^2$$
$$\omega L - \frac {1}{\omega C} = R$$
$$\omega^2 LC - \omega RC - 1 = 0$$
$$\omega^2 - \omega \frac {R}{L} - \frac {1}{LC} = 0$$
Which gives \$[2]\$ when applied to quadratic formula.
$$\omega^2 - \omega \frac {\omega_0}{Q} - \omega_0^2 = 0$$
Which gives \$[1]\$ when applied to quadratic formula.