+1904 I assume you want to solve for n which is what I will do.
If you mean \(f(n)=\sqrt{n+2}\) I will use this.
\(f(n)=\sqrt{n+2}\)
\({{f(n)}^{2}}={\sqrt{n+2}}^{2}\)
\({{f(n)}^{2}}=n+2\)
\({{f(n)}^{2}}-2=n+2-2\)
\({{f(n)}^{2}}-2=n+0\)
\({{f(n)}^{2}}-2=n\)
\(n={{f(n)}^{2}}-2\)
If you mean \(f(n)=\sqrt{n}+2\) I will use this
\(f(n)=\sqrt{n}+2\)
\(f(n)-2=\sqrt{n}+2-2\)
\(f(n)-2=\sqrt{n}+0\)
\(f(n)-2=\sqrt{n}\)
\({(f(n)-2)}^{2}={\sqrt{n}}^{2}\)
\({(f(n)-2)}^{2}=n\)
\({f(n)}^{2}-4f(n)+4=n\)
\(n={f(n)}^{2}-4f(n)+4\)
