5. Determine the smallest possible value of the expression $4 x^{2}+4 x y+4 y^{2}+12 x+8$. For which $x$ and $y$ will this expression have the smallest value?

Each task is scored out of 10 points.

The use of a pocket calculator or any reference materials is not allowed.
5. Let's rewrite the given expression in another form, using the formula for the square of a sum:

$$
\begin{aligned}
4 x^{2}+4 x y+4 y^{2}+12 x+8 & =x^{2}+4 x y+4 y^{2}+3 x^{2}+12 x+12-4 \\
& =(x+2 y)^{2}+3\left(x^{2}+4 x+4\right)-4 \\
& =(x+2 y)^{2}+3(x+2)^{2}-4
\end{aligned}
$$

Since the square of a number is always non-negative, we conclude that $(x+2 y)^{2} \geq 0$ and $(x+2)^{2} \geq 0$, so the minimum value is achieved when $(x+2 y)^{2}=(x+2)^{2}=0$, and it is equal to -4. The minimum value is achieved when $x+2 y=0$ and $x+2=0$. From the second equation, we get $x=-2$, and from the first, $y=-\frac{x}{2}=-\frac{-2}{2}=1$.