<s>[INST] If $x = y$, $p$ = what? It is given that,
$$
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p
$$
If $x=y$, $p$ =? 
 [/INST] $
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p
$
Apply componendo and dividendo,
$
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}+\sqrt{2x+3y}-\sqrt{2x-3y}}{\sqrt{2x+3y}+\sqrt{2x-3y}- \sqrt{2x+3y}+\sqrt{2x-3y}}=\frac{p+1}{p-1}
$
$
\frac{\sqrt{2x+3y}}{\sqrt{2x-3y}} = \frac{p+1}{p-1}
$
Put x=y
$
\frac{\sqrt{5y}}{\sqrt{-y}} = \frac{p+1}{p-1}
$
Squaring both sides,
$5.(p-1)^2=-1(p+1)^2$

Sorry I have little mistake here,
  $5p^2-10p+5 = -p^2-2p-2$

$5p^2-10p+5 = -p^2-2p-1$
$6p^2-8p+6=0$
$3p^2-4p+3=0$
$D = b^2 - 4ac$
= $(-4)^2 - 4.3.3$ = 16 - 36 = - 20
$ p = \frac{-b \pm \sqrt D}{2a}$  
= $\frac{-(-4) \pm \sqrt{-20}}{2.3}$
= $\frac{4 \pm \sqrt{20i^2}}{6}$
= $\frac{4 \pm 2i\sqrt5}{6}$
p = $\frac{2 + i\sqrt5}{3}, \frac{2 - i\sqrt5}{3}$

Edit- 

In above formula D is discriminat. It is used when we can't factorise equation using factorisation method.
Here's the link when you can't find root using factorisation use discrimat method.
</s>