From a paper I was reading, If:
$$w=\frac {1}{3}\left\{ \frac {-\dfrac {3}{16}\lambda^2}{1}+\frac {-\dfrac {3}{16}\lambda^2}{1}+\frac {-\dfrac {3}{16}\lambda^2}{1}+\frac {-\dfrac {11}{48}\lambda^2}{1}+\cdots\right\},$$
(which is a continued fraction) and "each numerator above is replaced by $-\dfrac{3}{16}\lambda^2$, then we obtain the approximation"
$$w \approx \frac {1}{12}\left( -2+\sqrt {4-3\lambda^2}\right).$$
I would greatly appreciate it if someone could explain how this step is calculated.
Replacing the numerators with $\tau:=-3\lambda^2/16$ gives the continued fraction the approximation
$$3w\approx \left\{ \dfrac {\tau}{1}+\dfrac {\tau}{1}+\dfrac {\tau}{1}+\cdots\right\}=\dfrac{\tau}{1+3w}\implies (3w)^2+3w=\tau\implies 3w=\frac{-1\pm \sqrt{1+4\tau}}{2}$$ We take the positive root since $\tau=0$ should give $w=0$ and we conclude that $$w\approx\frac{1}{6}\left(-1+\sqrt{1-\frac{3\lambda^2}{4}}\right)=\frac{1}{12}\left(2-\sqrt{4-3\lambda^2}\right).$$