If $$\sqrt{x}=\sqrt{3}-\sqrt{5}$$, then the value of $$x^2-16x+6$$ is:

Given,  $$\sqrt{x}=\sqrt{3}-\sqrt{5}$$

$$\Rightarrow$$  $$x=\left(\sqrt{3}-\sqrt{5}\right)^2$$

$$\Rightarrow$$  $$x=3+5-2\sqrt{15}$$

$$\Rightarrow$$  $$x=8-2\sqrt{15}$$ ...............(1)

$$\Rightarrow$$  $$x^2=\left(8-2\sqrt{15}\right)^2$$

$$\Rightarrow$$  $$x^2=64+60-32\sqrt{15}$$

$$\Rightarrow$$  $$x^2=124-32\sqrt{15}$$ ...........(2)

$$\therefore\ $$ $$x^2-16x+6=124-32\sqrt{15}-16\left(8-2\sqrt{15}\right)+6$$

$$=124-32\sqrt{15}-128+32\sqrt{15}+6$$

$$=130-128$$

$$=2$$

Hence, the correct answer is Option C