Let's determine the value of $\sqrt{3(7+4 \sqrt{3})}$ as the sum of two simple square roots!
We need to solve the equation

$$
\sqrt{3(7+4 \sqrt{3})}=\sqrt{x}+\sqrt{y}
$$

From this, we get $\quad 3(7+4 \sqrt{3})=x+y+2 \sqrt{x y}$

which means

$$
x+y=21 \quad \text{and} \quad 2 \sqrt{x y}=12 \sqrt{3} \quad \text{or} \quad x y=108 .
$$

Accordingly, $x$ and $y$ are the roots of the equation $\quad u^{2}-21 u+108=0$;

these roots are 9 and 12. Therefore,

$$
\sqrt{3(7+4 \sqrt{3})}=\sqrt{12}+\sqrt{9}=2 \sqrt{3}+3
$$

George Holló (Dobó István g. VI. o. Eger)