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see this questions is following:
first square both sides we get

\[(\sqrt{x}+2)^{2}p =2^{2}\]

p = 2sorry

ohk
now we know
(a+b)^2 = \[a^{2}+2ab +b ^{2}\]

\[hence, (\sqrt{x}^{2}+2(\sqrt{x}+2)+2^{2} = 4\]

now we get
\[x +2\sqrt{x}+4+4=4\]

\[x +2\sqrt{x} = 4\]

now we will solve (x=2)^2

\[x ^{2}+4x +4\]

i will continue my answer after some time i have to go seriously sorry

Thank you for all your help,