An algebra problem by Sabhrant Sachan

$\text {we start with A=5} \\ \implies A=\sqrt{25} \\ \implies A=\sqrt{1+24} \\ \implies A=\sqrt{1+4\times6} \\ \implies A=\sqrt{1+4\sqrt{36}} \\ \implies A=\sqrt{1+4\sqrt{1+5\times7}} \\ \implies A=\sqrt{1+4\sqrt{1+5\sqrt{49}}} \\ \implies A=\sqrt{1+4\sqrt{1+5\sqrt{1+6\times8}}} \\ \implies A=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{64}}}} \\ \implies A=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\cdots}}}} \\ \text{Hence} \boxed{A=x=5}$

This question is wrong as it can have infinite solutions...for example if we start with 3...with the same process we can have d result...but we have to continue with fractions...this infinite series was ramanujan's which can be started with any number...