$\underline{\textbf{Solution 1}}$ \begin{align} &\sqrt{7+4\sqrt{3}} + \sqrt{7-4\sqrt{3}}\\ &=\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}\\ &=\sqrt{(\sqrt{4})^2+(\sqrt{3})^2+2\cdot\sqrt{3}\cdot\sqrt{4}}+\sqrt{(\sqrt{4})^2+(\sqrt{3})^2-2\cdot\sqrt{3}\cdot\sqrt{4}}\\ &=\sqrt{(\sqrt{4}+\sqrt{3})^2} + \sqrt{(\sqrt{4}-\sqrt{3})^2}\\ &=\boxed{4} \end{align} Note: handling nested radical expressions is discussed in the book  Power Calculation by Example .

$\underline{\textbf{Solution 2}}$ Squaring the given expression gives \begin{align} &(7+4\sqrt{3})+(7-4\sqrt{3})+2\sqrt{7+4\sqrt{3}}\sqrt{7-4\sqrt{3}}\\ &=14 + 2\sqrt{49-48}\\ &=16 \end{align} Therefore, the original expression equals $4$.