What's x/y?

$\begin{aligned} x + \sqrt{y} + \sqrt{3} & = \sqrt{12 + 2\sqrt{15} + 4 \sqrt{5} + 4 \sqrt{3}} \quad \quad \small \color{#3D99F6}{\text{Squaring both sides}} \\ x^2 + y + 3 + 2x\sqrt{y} + 2\sqrt{\color{#3D99F6}{3y}} + \color{#D61F06}{2x}\sqrt{3} & = 12 + 2\sqrt{\color{#3D99F6}{15}} + 4 \sqrt{5} + \color{#D61F06}{4} \sqrt{3} \end{aligned}$

Equating coefficients, we have:

$\begin{cases} \color{#D61F06}{2x = 4} & \Rightarrow \color{#D61F06}{x = 2} \\ \color{#3D99F6}{3y = 15} & \Rightarrow \color{#3D99F6}{y = 5} \end{cases}$

$\Rightarrow \dfrac{x}{y} = \boxed{0.4}$

In such type of questions your main aim is to remove square root.

$\sqrt{12+2\sqrt{15}+4\sqrt{5}+4\sqrt{3}}$

$\sqrt{4+5+3+2(2\sqrt{5}+\sqrt{15}+2\sqrt{3})}$

$\sqrt{(2+\sqrt{5}+\sqrt{3})^2}$

$2+\sqrt{5}+\sqrt{3}$

On comparing with $x+\sqrt{y}+\sqrt{3}$ .

We get $x=2$ and $y=5$ .

$\therefore \dfrac{x}{y}=\dfrac{2}{5}=\boxed{0.4}$

Hint: transform LHS into a complete square