#### Answer

$f(5+\sqrt{2})=27+10\sqrt{2}$

#### Work Step by Step

Substituting $x$ with $(5+\sqrt{2})$ in $f(x)=x^2,$ then
\begin{array}{l}\require{cancel}
f(x)=x^2
\\\\
f(5+\sqrt{2})=(5+\sqrt{2})^2
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
f(5+\sqrt{2})=(5)^2+2(5)(\sqrt{2})+(\sqrt{2})^2
\\\\
f(5+\sqrt{2})=25+10\sqrt{2}+2
\\\\
f(5+\sqrt{2})=27+10\sqrt{2}
.\end{array}