Answer
$f(2-\sqrt{a})=4-4\sqrt{a}+a$
Work Step by Step
Substituting $x$ with $(2-\sqrt{a})$ in $f(x)=x^2,$ results to
\begin{array}{l}\require{cancel}
f(x)=x^2
\\\\
f(2-\sqrt{a})=(2-\sqrt{a})^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(2-\sqrt{a})=(2-\sqrt{a})^2
\\\\
f(2-\sqrt{a})=(2)^2-2(2)(\sqrt{a})+(\sqrt{a})^2
\\\\
f(2-\sqrt{a})=4-4\sqrt{a}+a
.\end{array}
