√(x+(√x-2)=2

thank you so much!

\(\sqrt{x+\sqrt{x-2}}=2 ===>> {(x+\sqrt{x-2})}^{1/2}=2 ==>>> (x+\sqrt{x-2})=4 ===>>>4-x =\sqrt{x-2}\)

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

\(16-8x+{x}^{2}=x-2====>>18-9x+{x}^{2}=0\)

Applying Bhaskara we gonna have x=6 or x=3

And 

\(\sqrt{3+\sqrt{3-2}}=2\)

X=3

Thanks I got the same answer!!