quadratic doubt

\hspace{-16}$Let $\mathbf{x=\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7.............\infty}}}}}$\\\\ $\mathbf{x=\sqrt{7+\sqrt{7-x}}\Leftrightarrow (x^2-7)=\sqrt{7-x}}$\\\\ $\mathbf{(x^2-7)^2=7-x}$\\\\\ $\mathbf{x^4+7^2-2.7.x^2=7-x}$\\\\ $\mathbf{x^4-2.7.x^2+x+(7^2-7)=0}$\\\\ Now Let $\mathbf{7=a}$. Then \\\\ $\mathbf{a^2-a.(2x^2+1)+(x^4+x)=0}$\\\\ So $\mathbf{a=\frac{(2x^2+1)\pm \sqrt{(2x^2+1)^2-4(x^4+x)}}{2}}$\\\\ $\mathbf{a=\frac{2x^2+1\pm (2x-1)}{2}}$\\\\ Now taking positive Sign::\\\\ $\mathbf{a=\frac{2x^2+1+2x-1}{2}=x^2+x}$\\\\ So $\mathbf{x^2+x=a=7}$\\\\ $\mathbf{x^2+x-7=0}$\\\\ $\mathbf{x=\frac{-1\pm \sqrt{29}}{2}}$\\\\ So $\mathbf{\boxed{\bold{x=\frac{-1 +\sqrt{29}}{2}\;,x=\frac{-1- \sqrt{29}}{2}}}}$\\\\\\

\hspace{-16}$Similarly Now taking Negative Sign::\\\\ $\mathbf{a=\frac{2x^2+1-2x+1}{2}=x^2-x+1}$\\\\ So $\mathbf{x^2-x+1=a=7}$\\\\ $\mathbf{x^2-x-6=0}$\\\\ $\mathbf{x=\frac{1\pm 5}{2}=-2\;,3}$\\\\ So $\mathbf{\boxed{\bold{x=-2\;,3}}}$\\\\ So $\mathbf{\boxed{\bold {x=3}}}$