$$
\sqrt{7+4 \sqrt{3}}+\sqrt{7-4 \sqrt{3}}=?
$$
I. solution. The question is about a positive number and its square:

$$
\begin{gathered}
(\sqrt{7+4 \sqrt{3}}+\sqrt{7-4 \sqrt{3}})^{2}=7+4 \sqrt{3}+2 \sqrt{(7+4 \sqrt{3})(7-4 \sqrt{3})}+ \\
+7-4 \sqrt{3}=14+2 \sqrt{49-48}=16
\end{gathered}
$$

thus, the value of the expression is $\sqrt{16}=4$.

II. solution. We can notice that $7+4 \sqrt{3}=(2+\sqrt{3})^{2}$, and $7-4 \sqrt{3}=(2-\sqrt{3})^{2}$. Since here $2>\sqrt{3}$, then

$$
\sqrt{7+4 \sqrt{3}}+\sqrt{7-4 \sqrt{3}}=(2+\sqrt{3})+(2-\sqrt{3})=4
$$