Manipulation of equation into quadratic form I am reading a paper for self study, and am trying to figure out how the authors can make the following claim.  They claim the following statements are equivalent:
$$\frac{c{\sqrt{d_1+d_2}}-\sqrt{d_1}z_1 -\delta(d_2) \sqrt{P_0P_1}}{\sqrt{d_1+d_2}} =  -z_{\beta}$$
and
$$(\delta^2p_1p_0)d_2^2 - d_2[(c+z_\beta)^2 -2\sqrt{d_1p_1p_0}z_1\delta] + d_1[z_1^2 - (c+z_{\beta})^2] =0$$
How do we rearrange the first equation so that it is a quadratic function of $d_2$?  Any hints would be appreciated!  I have tried multiplying both sides by $\sqrt{d_1+d_2}$ then squaring, and that didn't work.  Somehow I need to multiply $z_\beta$ by $c$, and I can't see when in the derivation that needs to happen.  Thank you!
Here is the paper for reference.

At first glance, it seems that the difficulty lies in the combination of terms that contain $d_2$ and $\sqrt{d_1+d_2}$ in the numerator, because if you simply square the expression, the two factors will mix, and you won't be able to get rid of the square root over $\sqrt{d_1 + d_2}$ in the mix term. However, note that the denominator itself is $\sqrt{d_1+d_2}$, so you can split off the term $c\sqrt{d_1+d_2}$ in the numerator, and then the left hand side becomes,
$$
\frac{c\sqrt{d_1+d_2} - z_1\sqrt{d_1} - d_2 \delta\sqrt{p_1p_0}}{\sqrt{d_1 + d_2}} = \frac{c\sqrt{d_1+d_2}}{\sqrt{d_1+d_2}} - \frac{z_1\sqrt{d_1}+d_2\delta\sqrt{p_1p_0}}{\sqrt{d_1+d_2}}\ .
$$
This allows you to rearrange the equation into
$$
\frac{z_1\sqrt{d_1} + d_2 \delta\sqrt{p_1p_0}}{\sqrt{d_1+d_2}} = c + z_\beta
$$
and further into
$$
z_1\sqrt{d_1} + d_2 \delta\sqrt{p_1p_0} = (c+z_\beta)\sqrt{d_1+d_2}\ .
$$
Now you can square both sides and rearrange it into a quadratic equation in $d_2$.
