$$\frac{\sqrt{3+x} + \sqrt{3-x}}{\sqrt{3+x} - \sqrt{3-x}}$$ = 2 then x is equal to

Let's say $$\sqrt{3+x}$$ = a
and $$\sqrt{3-x}$$ = b
hence $$\frac{a+b}{a-b}$$ = 2
Now let's say $$\frac{a}{b}$$ = $$y$$
According to ratio proportionate rule  $$\frac{a+b}{a-b}$$ =  $$\frac{y+1}{y-1}$$
After putting value and solving for $$y$$, its value will be 3
hence $$\frac{a}{b}$$ = 3
now putting a = $$\sqrt{3+x}$$ and $$\sqrt{3-x}$$ = b and solving for $$x$$
we will get $$x$$=$$\frac{12}{5}$$