Simplify the radical $\sqrt{x-\sqrt{x+\sqrt{x-...}}}$
Perhaps it is more instructive to consider instead the following: let $$y = \sqrt{x - \sqrt{x + \sqrt{x - \sqrt{x + \cdots}}}}, \\ z = \sqrt{x + \sqrt{x - \sqrt{x + \sqrt{x - \cdots}}}},$$ so that if $y$ and $z$ exist, they satisfy the system $$y = \sqrt{x - z}, \\ z = \sqrt{x + y},$$ or $$y^2 = x - z, \\ z^2 = x + y.$$ Consequently $$0 = z^2 - y^2 - y - z = (z-y-1)(y+z).$$ It follows that either $z = -y$ or $z = 1 + y$. The first case is impossible for $x \in \mathbb R$ since by convention we take the positive square root, so both $y, z > 0$. In the second case, we can substitute back into the first equation to obtain $y^2 = x - (1+y)$, hence $$y = \frac{-1 + \sqrt{4x-3}}{2},$$ where again, we discard the negative root.
So far, what we have shown is that if such a nested radical for $y$ converges, it must converge to this value. It is not at all obvious from the above whether a given choice of $x$ results in a real-valued $y$, for any meaningful definition of $y$ must be as the limit of the sequence $$y = \lim_{n \to \infty} y_n, \\ y_n = \underbrace{\sqrt{x - \sqrt{x + \sqrt{x - \cdots \pm \sqrt{x}}}}}_{n \text{ radicals}},$$ and although the choice $x = 1$ appears at first glance permissible, we quickly run into problems; $y_3 = \sqrt{1 - \sqrt{1 + \sqrt{1}}} \ne \mathbb R$. In particular, we need $x$ to satisfy the relationship $$x \ge \sqrt{x + \sqrt{x}},$$ which leads to the cubic $x^3 - 2x^2 + x - 1$ with unique real root $$x = \frac{1}{3} \left(2+\sqrt[3]{\frac{25-3 \sqrt{69}}{2}}+\sqrt[3]{\frac{25+3 \sqrt{69}}{2}}\right) \approx 1.7548776662466927600\ldots.$$ However, any such $x$ meeting this condition will lead to a convergent sequence. The idea is to show that $|y_{n+2} - y| < |y_n - y|$ for all $n \ge 1$; then since $\lim y_n$ has at most one unique limiting value as established above, the result follows.
Consider the final relation you have obtained as a quadratic equation in $x$,i.e: $$x^2-(2y^2+1)x+y^4-y=0$$ Solving the above gives you $$x=y^2+y+1 \text{ or } x=y^2-y$$ Individually solve these the quadratics in $y$ to obtain the four solutions you got from Wolfram Alpha.
Note that :
$$(x-y^2)^2 = x+y \implies (x-y^2)^2 - y^2 = x+y-y^2 \implies (x-y^2-y)(x-y^2+y) = x-y^2+y \\ \implies \boxed{(x-y^2+y)(x-y^2-y-1) = 0}$$
So either one is correct.
Note : The problem is that one is still not sure when the radical above converges i.e. what is the set of all $x$ for which $\sqrt{x + \sqrt{x-\sqrt{x+...}}}$ forms a convergent sequence.