Finding upper cutoff and lower cutoff frequencies on notch circuit
If it's about calculating the bandwidth, it would be better to calculate the transfer function instead of relying on pre-made formulas:
$$\begin{align}
Z_{ser}(s)&amp;=R_{ser}+sL\tag{1} \\
Z_{par}(s)&amp;=\frac{1}{\frac{1}{Z_{ser}}+sC}\tag{2} \\
H(s)&amp;=\frac{R}{R+Z_{par}(s)} \\
&amp;\Rightarrow \\
H(s)&amp;=\frac{RLCs^2+RR_{ser}Cs+R}{RLCs^2+(RR_{ser}C+L)s+R_{ser}+R} \\
{}&amp;=\frac{s^2+\frac{R_{ser}}{L}s+\frac{1}{LC}}{s^2+\left(\frac{R_{ser}}{L}+\frac{1}{RC}\right)s+\frac{R_{ser}+R}{R}\frac{1}{LC}}\tag{3}
\end{align}$$
This gives a transfer function in the generic form:
$$G(s)=K\frac{s^2+\frac{\omega_z}{Q_z}s+\omega_z^2}{s^2+\frac{\omega_p}{Q_p}s+\omega_p^2}=K\frac{s^2+BW_zs+\omega_z^2}{s^2+BW_ps+\omega_p^2}\tag{4}$$
From \$(3)\$ and \$(4)\$ the center frequency and the bandwidths can be extracted:
$$\begin{align}
K=H(0)&amp;=\frac{R}{R+R_{ser}}\tag{5} \\
\omega_p=\omega_z&amp;=\frac{1}{LC}\tag{6} \\
BW_z&amp;=\frac{R_{ser}}{L}\tag{7} \\
BW_p&amp;=\frac{R_{ser}}{L}+\frac{1}{RC}\tag{8} \\
\end{align}$$
Which shows that the formulas in your post discard the \$R_{ser}\$ element and, for small enough values it can be discarded. In this case: \$25\;\Omega\$ and \$15\;\text{k}\Omega\$, so it's fine. To verify use a calculator or simulator to compare the circuit with the Laplace expression:

The RLC version overlaps with the Laplace expression, so it's a match. The measured quantities are the frequencies, f1 and f2, the passband bandwidth bw, and the center frequency, fc, against the calculated bwp and f0. The numbers come very close, save precision, parasitics, etc.

Given your formulation I'm not sure how to interpret your question, but I'll add this, to be sure. In case what you want is to find independent formulas for \$f_L\$ and \$f_H\$, then all you have to do is use the following system of equations:
$$\left\{
\begin{aligned}
f_Hf_L&amp;=\frac{1}{LC} \\
f_H-f_L&amp;=\frac{R_{ser}}{L}+\frac{1}{RC} \\
\end{aligned}
\right.$$
which results in a 2nd order equations with the two valid roots:
$$f_L=\frac{\sqrt{(RR_{ser}C)^2+2RLC(R_{ser}+2R)+L^2}-RR_{ser}C-L}{2RLC} \\
f_H=\frac{\sqrt{(RR_{ser}C)^2+2RLC(R_{ser}+2R)+L^2}+RR_{ser}C+L}{2RLC}$$
And the result of the .meas scripts are:
f1: freq=(2.36937e+006,0) at 2.36937e+006
f2: freq=(2.66826e+006,0) at 2.66826e+006
f_lo: (sqrt(c**2*r**2*rs**2+2*c*l*r*rs+4*c*l*r**2+l**2)-c*r*rs-l)/(2*c*l*r)/2/pi=(2.36856e+006,0)
f_hi: (sqrt(c**2*r**2*rs**2+2*c*l*r*rs+4*c*l*r**2+l**2)+c*r*rs+l)/(2*c*l*r)/2/pi=(2.6736e+006,0)