Solution
Let $$\sqrt{\ 6+\sqrt{\ 6+\sqrt{\ 6+........}}} = x$$
Therefore, $$\sqrt{\ 6+x} = x$$
$$6+x = x^{2}$$
$$x^{2}-x-6 = 0$$
$$x^2-3x+2x-6=0$$
$$x\left(x-3\right)+2\left(x-3\right)=0$$
$$\left(x-3\right)\left(x+2\right)=0$$
$$x=3,\ x=-2$$
Hence, $$\sqrt{\ 6+\sqrt{\ 6+\sqrt{\ 6+........}}}=\ 3$$
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