Four square roots

$x=\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-x}}}}\implies x=\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+\cdots}}}}}$ Now $x=\sqrt{7+\sqrt{7-x}} \implies x^2= 7+\sqrt{7-x} \implies( x^2-7)^2=7-x$ $\implies x^4-14x^2+x+42=0 \implies(x-3)(x+2)(x^2+x-7)=0$ $x= -2$ (Not possible) , $x=\dfrac{-1+\sqrt{29}}{2}$ (Not possible) $\dfrac{\sqrt{29}-1}{2} ≠ \sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-(\dfrac{\sqrt{29}-1}{2})}}}}$

Solution is $\color{#20A900}\boxed{x=3}$