We have
\[x=\sqrt{7+4\sqrt 3}+\sqrt{7-4\sqrt 3}\]
Let's multiply and divide this by \(\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}\)
we have
\[x=\sqrt{7+4\sqrt 3}+\sqrt{7-4\sqrt 3} \times \frac{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}\]
We get
\[x=\frac{{(\sqrt{7+4\sqrt 3})^2 }-(\sqrt{(7-4\sqrt 3})^2 }{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}\]
we get
\[x=\frac{7+4\sqrt 3-(7-4 \sqrt 3)}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}\]
so we get
\[x=\frac{8\sqrt 3}{\sqrt{7+4\sqrt 3}-\sqrt{7-4\sqrt 3}}\]
We have
\[x=\sqrt {7+4 \sqrt 3}+\sqrt {7-4 \sqrt 3}\]
Let's square both the sides
we get
\[x^2=(\sqrt {7+4 \sqrt 3})^2+(\sqrt {7-4 \sqrt 3})^2+2\times (\sqrt {7+4 \sqrt 3})\times
(\sqrt {7-4 \sqrt 3})\]
we get
\[x^2=7+4\sqrt 3+7-4\sqrt 3+2 \sqrt{(7^2-(4\sqrt 3)^2}\]
we get
\[x^2=14+2\times \sqrt {49-16\times 3}\]
we get
\[x^2=14+2\times \sqrt 1\]
we get
\[x^2=14+2=16\]
so
\[x=4\]