If $x = y$, $p$ = what? It is given that,
$$
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p
$$
If $x=y$, $p$ =? 
 A: We have: $\dfrac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p$
Let's replace $x$ with $y$:
$\Rightarrow p=\dfrac{\sqrt{2y+3y}+\sqrt{2y-3y}}{\sqrt{2y+3y}-\sqrt{2y-3y}}$
$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{-y}}{\sqrt{5y}-\sqrt{-y}}$
$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}-\sqrt{y}\hspace{0.5 mm}i}$
Then, let's multiply this fraction by the complex conjugate of its denominator:
$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}-\sqrt{y}\hspace{0.5 mm}i}\cdot{\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}}$
$\hspace{9 mm} =\dfrac{\big(\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i\big)^{2}}{\big(\sqrt{5y}\big)^{2}-\big(\sqrt{y}\hspace{0.5 mm}i\big)6{2}}$
$\hspace{9 mm} =\dfrac{5y+2\cdot{\sqrt{5y}}\cdot{\sqrt{y}\hspace{0.5 mm}i}+y\cdot{i^{2}}}{5y-y\cdot{i^{2}}}$
$\hspace{9 mm} =\dfrac{5y+2{\sqrt{5y^{2}}\hspace{0.5 mm}i}-y}{5y+y}$
$\hspace{9 mm} =\dfrac{4y+2{\sqrt{5y^{2}}\hspace{0.5 mm}i}}{6y}$
$\hspace{9 mm} =\dfrac{2y+2\sqrt{y^{2}}\sqrt{5}\hspace{0.5 mm}i}{3y}$
$\hspace{9 mm} =\dfrac{2y+2y\sqrt{5}\hspace{0.5 mm}i}{3y}$
$\hspace{9 mm} =\dfrac{2+2\sqrt{5}\hspace{0.5 mm}i}{3}$; $y\neq{0}$
A: $
\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.
A: Since $x=y$, let's call them both $a$.
Then substituting, we have:
$$
p=\frac{\sqrt{5a}+\sqrt{-a}}{\sqrt{5a}-\sqrt{-a}}
$$
If we multiply the top and bottom by the conjugate, $\sqrt{5a}-\sqrt{-a}$, we get:
$$
p=\frac{5a-(-a)}{5a-2\cdot\sqrt{5a}\cdot\sqrt{-a}+(-a)}
$$$$
p=\frac{6a}{4a-2\cdot\sqrt{-5a^2}}=\frac{6a}{4a-2\cdot a\cdot\sqrt{-5}}
$$$$
p=\frac6{4-2\cdot\sqrt{5}\cdot\sqrt{-1}}=\frac3{2-\sqrt{5}\cdot i} [if a\neq 0]
$$
This can be further simplified, but from here, a calculator can be used to solve the complex fraction.
