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Solution
$$x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$$
=> $$x = \sqrt{4+\sqrt{4-x}}$$
=> $$x^2 = 4 + \sqrt{4-x}$$
=>$$x^4 + 16 - 8x^2 = 4 - x$$
=> $$x^4 - 8x^2 + x +12 = 0$$
On substituting options, we can see that option C satisfies the equation.
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