Solution
we need to find value of $$\sqrt{6+\sqrt{6+\sqrt{6+...}}}$$
let $$\sqrt{6+\sqrt{6+\sqrt{6+...}}}$$ = x
here x = $$\sqrt{6+x}$$
on squaring both sides
$$x^2 - x - 6$$ = 0
x = 3 , x = -2
here -2 will be rejected as square root can not give negative value and hence x = 3
$$\sqrt{6+\sqrt{6+\sqrt{6+...}}}$$ = 3
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