Let $\sqrt{3+\sqrt{3-\sqrt{3+\sqrt{3-\sqrt{...}}}}}=P$

and $\sqrt{3-\sqrt{3+\sqrt{3-\sqrt{3+\sqrt{...}}}}}=Q$

So $\sqrt{3+Q}=P .......... (1)$

$\sqrt{3-P}=Q ........... (2)$

Square both sides Subtract ep. (2) from eq(1) to get

$P^{2}-Q^{2}=P+Q$

$Q=P-1$

Substitute the value of $Q$ in eq. (1) to get

$P^{2}-P-2=0$

$(P-2)(P+1)=0$

So $P=\boxed{2}$

This solution is completely inspired from @Satvik Golechha .

Correct answer is NOT only 2, there could be another possible solution you could have forgotten.

Let $x=\sqrt{3+\sqrt{3-\sqrt{3+\sqrt{3-\sqrt{\dots}}}}}$

Then $x=\sqrt{3+\sqrt{3-x}}$ . Taking squares in both sides, you'll get $x^2=3+\sqrt{3-x}$ . Moving 3 to the left side, and taking squares once again $(x^2-3)^2=3-x$ and therefore $p(x)=x^4-6x^2+x+6=0$

Factoring, we have $p(x)=(x+1)(x-2)(x^2+x-3)\\=(x+1)(x-2)(x-\frac{\sqrt{13}-1}{2})(x-\frac{-\sqrt{13}-1}{2})$

So, all the possible solutions are, rounding to 2 decimal places, $x=-1,2,1.30,-2.30$

Discarding negative solutions (x is the result of a square root, so the convention is to choose the positive sign for it), we have $x=2$ , BUT $x=1.30277563...$ as well...

Well, to finish, to discard $x=1.30277563...$ , we realize these are solution of a radical equation, so we MUST test the solution in the original expression, BOTH of them.

For $x=2$ , it's ok: $x=?\sqrt{3+\sqrt{3-2}}=\sqrt{3+\sqrt{1}}=\sqrt{3+1}=\sqrt{4}=2$

But, for $x=1.30...$ , $\sqrt{3+\sqrt{3-1.30}}=\sqrt{3+\sqrt{1.7}}=\sqrt{3+1.3\dots}=\sqrt{4\dots}=2\dots \neq x=1.30 ...$

So, at the end, the only solution is $\boxed{x=2}$

$x=\sqrt{3+\sqrt{3-x}}\implies (x^2-3)^2=3-x$ Put $x=2$

$\text{Game Over!}$

Horner method work fine

tbhjufygjuh