Solve in $\mathbb{R}$, $\sqrt{x+1+\sqrt{x+1+\sqrt{x+3}}}=\sqrt{x+3}$

Solution:
Let \({u}^{2}=x+3\) $\Rightarrow$ $\sqrt{x+1+\sqrt{x+1+\sqrt{{u}^{2}}}}=\sqrt{{u}^{2}}$ $\Rightarrow$ $\sqrt{x+1+\sqrt{x+1+u}}=u$ Now Squaring both sides to get :
$ x+1+\sqrt{x+1+u}={u}^{2} $ $\Rightarrow$ $\sqrt{x+1+u}={u}^{2}-x-1 $ but ${u}^{2}=x+3 $ hence $\sqrt{x+1+u}=x+3-x-1 $ thus $\sqrt{x+1+u}=2 $ again Squaring both sides to get $ x+1+u=4 $ $\Rightarrow$ $x+u=3$ but $ u=\sqrt{x+3} $ thus $x+\sqrt{x+3}=3$ Squaring both sides $x+3={(3-x)}^{2}=9-6x+{x}^{2}$ thus $ 7x-6-{x}^{2}=0$ $\Rightarrow$ $-({x}^{2}-7x+6)=0$ so ${{x}^{2}}-7x+{{\left( \frac{7}{2} \right)}^{2}}-{{\left( \frac{7}{2} \right)}^{2}}+6=0$ $\Rightarrow$ ${\left(x-\frac{7}{2}\right)}^{2}=-6+\frac{49}{4} $ $\Rightarrow$ ${{\left( x-\frac{7}{2} \right)}^{2}}=\frac{-24+49}{4}=\frac{25}{4}$ $\Rightarrow$ $x-\frac{7}{2}=\pm \frac{5}{2}$ 
$\Rightarrow $ $x=\frac{7\pm 5}{2}$ $\Rightarrow$ $x=\frac{12}{2}=6\,\,or\,x=\frac{2}{2}=1\Leftrightarrow x=1 \,\,or\,\, 6 $