Let's find the minimum of the expression $6 t^{2}+3 s^{2}-4 s t-8 t+6 s+5$.

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Let's find the minimum of the expression $6 t^{2}+3 s^{2}-4 s t-8 t+6 s+5$.
Let the two-variable polynomial given in the problem be denoted by $f(t, s)$. This polynomial is quadratic in both variables; let's arrange it according to the powers of one variable, for example, $t$:

$$
f(t, s)=6 t^{2}-(4 s+8) t+3 s^{2}+6 s+5
$$

Transform this into a perfect square with respect to $t$:

$$
f(t, s)=6\left(t-\frac{s+2}{3}\right)^{2}+\frac{7}{3} s^{2}+\frac{10}{3} s+\frac{7}{3}
$$

From this, it is clear that for a fixed value of $s$, the minimum value of $f(t, s)$ is $\frac{7}{3} s^{2}+\frac{10}{3} s+\frac{7}{3}$, which is achieved when $t=\frac{s+2}{3}$. The minimum value of this expression can also be obtained by completing the square:

$$
\frac{7}{3} s^{2}+\frac{10}{3} s+\frac{7}{3}=\frac{7}{3}\left(s+\frac{5}{7}\right)^{2}+\frac{8}{7}
$$

The smallest minimum value is therefore $\frac{8}{7}$, which is achieved when $s=-\frac{5}{7}$. In this case, $t=\frac{s+2}{3}=\frac{3}{7}$.

Thus, the minimum value of $f(s, t)$ is $\frac{8}{7}$, which is achieved when $t=\frac{3}{7}$ and $s=-\frac{5}{7}$.

Barcza Zsófia (Bonyhád, Petőfi Sándor Evangelical Gymnasium, 9th grade)

Remarks. 1. We can achieve the same result if we first complete the square with respect to $s$, and then with respect to $t$. In this case, we get the form:

$$
f(s, t)=3\left(s-\frac{2}{3} t+1\right)^{2}+\frac{14}{3}\left(t-\frac{3}{7}\right)^{2}+\frac{8}{7}
$$

From this, it is also clear that the minimum value is $\frac{8}{7}$, which is achieved when $t-\frac{3}{7}=s-\frac{2}{3} t=0$, i.e., $t=\frac{3}{7}$ and $s=-\frac{5}{7}$.

2. From the form $f(s, t)=6\left(t-\frac{s+2}{3}\right)^{2}+\frac{7}{3}\left(s+\frac{5}{7}\right)^{2}+\frac{8}{7}$, it is also clear that in the $(s, t)$ coordinate system, $f(s, t)=c$ is the equation of an ellipse if $c>\frac{8}{7}$. If $c=\frac{8}{7}$, then the ellipse shrinks to a single point, whose coordinates are the values that yield the minimum: $t=\frac{3}{7}$ and $s=-\frac{5}{7}$. Our result also implies that if $c<\frac{8}{7}$, there is no point whose $(s, t)$ coordinates satisfy $f(s, t)=c$.