= Solution
Divide the equilibrium equation by $q>0$ to get $bq^2-q+a=0$. Hence both positive branches, when $0<4ab<1$, are
$$
q_\pm=\frac{1\pm\sqrt{1-4ab}}{2b}=\frac{C_uSL_u}{2}\left[1\pm\sqrt{1-4\frac{L_\rho C_\rho^2}{L_uC_u^2}\mathrm{Ri}}\right].
$$
At equilibrium, $\mathrm{Ri}_f=a/q=1-bq$. Therefore the larger-speed branch $q_+$ gives the \b[printed flux Richardson number]
$$
\boxed{\mathrm{Ri}_f=\frac12\left[1-\sqrt{1-4\frac{L_\rho C_\rho^2}{L_uC_u^2}\mathrm{Ri}}\right],}
$$
while the smaller-speed branch gives the same formula with a plus sign before the square root. Solving the quadratic alone does not discard that second branch.
The larger-speed branch is selected by local kinetic-energy stability. If the same closures are used in the local energy evolution, $\tfrac12d(q^2)/dt=\mathcal P-B-\epsilon$. For $q>0$, divide by $q$ to obtain
$$
\dot q=C_uSq-C_\rho^2L_\rho N^2-q^2/L_u.
$$
Its linearization at $q_\pm$ has derivative $C_uS-2q_\pm/L_u=\mp C_uS\sqrt{1-4ab}$. The upper branch is therefore stable and the lower unstable in this local closure model.
The discriminant gives the \b[critical gradient Richardson number]
$$
\boxed{\mathrm{Ri}_c=\frac{L_uC_u^2}{4L_\rho C_\rho^2}.}
$$
At criticality the branches meet and $\mathrm{Ri}_f=1/2$. At zero stratification the stable positive solution is $q=C_uSL_u$ with $\mathrm{Ri}_f=0$; the lower algebraic root is $q=0$. This closure-dependent critical value is not automatically the inviscid $1/4$ criterion of the <Miles–Howard theorem>.
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