Just Think of Quadratics

$\sqrt{4+\sqrt{4-x}} = x$

$(\sqrt{4+\sqrt{4-x}})^2 = (x)^2$

$4+\sqrt{4-x} = x^2$

$\sqrt{4-x} = x^2-4$

$(\sqrt{4-x})^2 = (x^2 - 4)^2$

$4-x = (x^2)^2 - 2x^2(4) + 4^2$

$4-x = x^4 - 8x^2 + 16$

$0 = x^4 - 8x^2 +x + 12$

We get $x= \dfrac{1+\sqrt{13}}{2}, x= \dfrac{1- \sqrt{17}}{2}, x= -\dfrac{1- \sqrt{13}}{2}, x= -\dfrac{1+\sqrt{17}}{2}$

Out of the four, $x = \dfrac{1+\sqrt{13}}{2}$ satisfy the equation.

Let $y = \sqrt{4-x}$ , so that the equation becomes $\sqrt{4+y} = x$ where both $x$ and $y$ are required to be non-negative, being square roots.

Squaring both sides of these equations gives

$\begin{aligned} y^2 &= 4 - x, \quad \quad (1)\\ x^2 &= 4 + y. \quad \quad (2) \end{aligned}$

Subtracting $(1)$ from $(2)$ gives $x^2 - y^2 = y+x$ or $(x+y)(x-y-1) = 0$ . Hence $y = -x$ or $y = x-1$ .

The first of these options is not possible since $x$ and $y$ are both non-negative and $x = y = 0$ does not satisfy the equations. The second case $y = x-1$ leads upon substitution in $(1)$ to $(x-1)^2 = 4-x$ from which $x^2 - x - 3 = 0$ . This has the solutions $x = \frac{1 \pm \sqrt{13}}{2}$ . Among these only the solution $\boxed{x = \frac{1}{2}(1 + \sqrt{13})}$ is non-negative and satisfies the original equation as

$\begin{aligned}\sqrt{4 + \sqrt{4-x}} &= \sqrt{4 + \sqrt{4 - \frac{1 + \sqrt{13}}{2}}} \\ &= \sqrt{4 + \sqrt{\frac{14 - 2\sqrt{13}}{4}}}\\ &= \sqrt{4 + \frac{\sqrt{13}-1}{2}}\\ &= \sqrt{\frac{14 + 2\sqrt{13}}{4}}\\ &= \frac{1 + \sqrt{13}}{2}\\ &= x.\end{aligned}$

Note that for general $c \geq 1$ we can repeat the above argument and show that $x = \frac{1}{2}(1 + \sqrt{4c-3})$ . (We require $c \geq 1$ to ensure that $y = x-1 \geq 0$ .)