Problem 2. Determine the angles of a triangle whose sides $a, b$ and $c$ satisfy the equation

$$
a-4 \sqrt{b c}+2 b=2 \sqrt{a c}-3 c
$$
Solution. We will write the equation in the form

$$
\begin{aligned}
& a-2 \sqrt{a c}+c+2 c-4 \sqrt{b c}+2 b=0 \\
& (\sqrt{a})^{2}-2 \sqrt{a} \sqrt{c}+(\sqrt{c})^{2}+2\left[(\sqrt{b})^{2}-2 \sqrt{b} \sqrt{c}+(\sqrt{c})^{2}\right]=0 \\
& (\sqrt{a}-\sqrt{c})^{2}+2(\sqrt{b}-\sqrt{c})^{2}=0
\end{aligned}
$$

The sum of two non-negative numbers is zero only if each of them is zero. Therefore,

$$
\begin{aligned}
& (\sqrt{a}-\sqrt{c})^{2}=0 \\
& (\sqrt{b}-\sqrt{c})^{2}=0
\end{aligned}
$$

The square of a number is zero only if the number itself is zero. Therefore, from (1) and (2) we get $\sqrt{a}-\sqrt{c}=0$ and $\sqrt{b}-\sqrt{c}=0$, i.e., $\sqrt{a}=\sqrt{c}$ and $\sqrt{b}=\sqrt{c}$. Due to these equalities, we have $a=c=b$. Thus, the triangle is equilateral, and its angles are equal to each other.