Solve the following equation if $x$ and $y$ are integers:

$$
5 x^{2}+5 x y+5 y^{2}=7 x+14 y
$$
Let's multiply both sides of the equation by 20. Thus we get

$$
100 x^{2}+100 x y+100 y^{2}=140(x+2 y)
$$

from which, after rearrangement,

$$
25(x+2 y)^{2}-2 \cdot 14[5(x+2 y)]+75 x^{2}=0
$$

that is,

$$
[5(x+2 y)-14]^{2}+75 x^{2}=14^{2}
$$

follows.

According to (1), $75 x^{2} \leqq 14^{2}$, since the other term of the sum is a square and thus non-negative. Since $x$ is an integer, this is only possible if $|x| \leqq 1$, that is, the possible values of $x$ are $-1,0$ and 1.

After substitution, the quadratic equations obtained for $y$ each have one integer root and the other is not, and since our steps are reversible, the equation has three solutions in the integers. These are

$$
\begin{array}{ll}
x_{1}=-1, & y_{1}=3 \\
x_{2}=0, & y_{2}=0 \\
x_{3}=1, & y_{3}=2
\end{array}
$$

Remark. We can also arrive at the solution in a different way, if we rearrange the given two-variable polynomial according to the powers of one of the variables, say $y$, and write down the discriminant of the resulting parametric equation. Since we are looking for integer solutions of the equation, $D=196-75 x^{2}$ must be a perfect square, which is only true if $x=-1,0$ or 1, and among the corresponding $y$ values, one in each of the three cases turns out to be an integer.