2. Determine the numbers $x, y$ and $z$ for which

$$
4 x^{2}+9 y^{2}+16 z^{2}-4 x-6 y-8 z+3=0
$$
Solution. The given equation is equivalent to the equation

$$
(2 x-1)^{2}+(3 y-1)^{2}+(4 z-1)^{2}=0
$$

Since the square of a real number is greater than or equal to zero, and the sum of non-negative numbers is equal to zero only if they are equal to zero, we get $2 x-1=3 y-1=4 z-1=0$, from which we find $x=\frac{1}{2}, y=\frac{1}{3}$ and $z=\frac{1}{4}$.