algebra precalculus - how can one solve for $x$, $x =sqrt{2+sqrt{2+sqrt{2cdots }}}$

how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$

we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$

now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$