#### Answer

$a=2$

#### Work Step by Step

Replacing $x$ with $a$ in $f(x)=\sqrt{x+2}+x$ results to
\begin{array}{l}\require{cancel}
f(x)=\sqrt{x+2}+x
\\\\
f(a)=\sqrt{a+2}+a
.\end{array}
Substituting $f(a)$ with $4$ and then isolating the radical expression result to
\begin{array}{l}\require{cancel}
f(a)=\sqrt{a+2}+a
\\\\
4=\sqrt{a+2}+a
\\\\
4-a=\sqrt{a+2}
.\end{array}
Squaring both sides, the equation above is equivalent to
\begin{array}{l}\require{cancel}
4-a=\sqrt{a+2}
\\\\
(4-a)^2=\left( \sqrt{a+2} \right)^2
\\\\
(4)^2-2(4)(a)+(a)^2=a+2
\\\\
16-8a+a^2=a+2
\\\\
a^2-8a-a+16-2=0
\\\\
a^2-9+14=0
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
a^2-9+14=0
\\\\
(a-7)(a-2)=0
.\end{array}
Equating each factor to zero and then isolating the variable, the solutions are $
a=\{2,7 \}$
If $a=7,$ then
\begin{array}{l}\require{cancel}
4=\sqrt{a+2}+a
\\\\
4=\sqrt{7+2}+7
\\\\
4=\sqrt{9}+7
\\\\
4=3+7
\\\\
4=10
\text{ (FALSE)}
.\end{array}
Hence, only $
a=2
$ satisfies the original equation.