How to calculate quantile function for Birnbaum–Saunders distribution? According to wikipedia the quantile function of for Birnbaum–Saunders distribution, $ G(p)$, depends on the quantile function of the standard normal distribution.
For example, in the paper https://arxiv.org/pdf/1805.06730.pdf it comes "evidently" on page 8, but I don't understand clear why it is almost the same as equation for $T$. How it was found?
 A: I use the notation in the paper you've linked (in case the link in the question dies, it is Birnbaum-Saunders Distribution: A Review
of Models, Analysis and Applications by N. Balakrishnan and Debasis Kundu).
We have that if $T \sim \text{BS}(\alpha, \beta)$ that
$$F_{T}(t) = \Phi\left[\dfrac{1}{\alpha}\left\{\left(\dfrac{t}{\beta}\right)^{1/2} - \left(\dfrac{\beta}{t}\right)^{1/2}  \right\}\right]\text{, } \quad t > 0\text{, } \alpha > 0\text{, } \beta > 0\text{.}$$
The $q$th quantile, by definition, is the value $t_q$ (which we assume is $>0$) satisfying
$$F_{T}(t_q) = \Phi\left[\dfrac{1}{\alpha}\left\{\left(\dfrac{t_q}{\beta}\right)^{1/2} - \left(\dfrac{\beta}{t_q}\right)^{1/2}  \right\}\right] = q\text{.}$$
As $\Phi$ is invertible, we obtain
$$\dfrac{1}{\alpha}\left\{\left(\dfrac{t_q}{\beta}\right)^{1/2} - \left(\dfrac{\beta}{t_q}\right)^{1/2}  \right\} = \Phi^{-1}(q) := z_q \tag{1}$$
because $\Phi^{-1}(q)$ is the $q$th quantile of a standard normal random variable. To make the algebra easier, we first observe that
$$\left(\dfrac{t_q}{\beta}\right)^{1/2} - \left(\dfrac{\beta}{t_q}\right)^{1/2} = \dfrac{t_q^{1/2}}{\beta^{1/2}} - \dfrac{\beta^{1/2}}{t_q^{1/2}} = \dfrac{t_q - \beta}{t_q^{1/2}\beta^{1/2}} = \dfrac{1}{\sqrt{\beta}}\left(\dfrac{t_q - \beta}{\sqrt{t_q}}\right)\text{.}$$
From $(1)$ and our work above, we obtain that
$$\alpha \sqrt{\beta} z_q = \dfrac{t_q - \beta}{\sqrt{t_q}} \implies t_q - \alpha\sqrt{\beta}z_q\sqrt{t_q} - \beta  = 0\text{.}$$
Let $u = \sqrt{t_q}$, then we have the quadratic
$$u^2 - \alpha\sqrt{\beta}z_qu - \beta = 0\text{.}$$
It follows from the quadratic formula that
$$\begin{align}
u &= \sqrt{t_q} \\
&= \dfrac{\alpha\sqrt{\beta}z_q \pm \sqrt{\alpha^2\beta z_q^2 - 4(1)(-\beta)}}{2} \\
&= \dfrac{\sqrt{\beta}}{2}\left(\alpha z_q \pm \sqrt{\alpha^2 z_q^2 + 4} \right)  \\
&= \dfrac{\sqrt{\beta}}{2}\left(\alpha z_q \pm \sqrt{(\alpha z_q)^2 + 4} \right)\text{.}
\end{align}$$
Now we observe that
$$\sqrt{(\alpha z_q)^2 + 4} > \sqrt{(\alpha z_q)^2} = \alpha |z_q| \geq \alpha z_q$$
hence
$$ \alpha z_q - \sqrt{(\alpha z_q)^2 + 4} < 0$$
so, with the condition that $\sqrt{t_q} \geq 0$, we obtain the unique solution
$$\sqrt{t_q} = \dfrac{\sqrt{\beta}}{2}\left(\alpha z_q + \sqrt{(\alpha z_q)^2 + 4} \right)$$
or
$$t_q = \dfrac{\beta}{4}\left(\alpha z_q + \sqrt{(\alpha z_q)^2 + 4} \right)^2$$
This matches equation $(8)$ in the paper.
