4. Solve the equation: $3 x^{2}+14 y^{2}-12 x y+6 x-20 y+11=0$.
Answer: $\boldsymbol{x}=3, \boldsymbol{y}=2$

Solution.

Consider the equation as a quadratic in $\boldsymbol{x}$, with $\boldsymbol{y}$ treated as a parameter. The equation $3 x^{2}-6 x(2 y-1)+14 y^{2}-20 y+11=0$ has real roots if and only if its discriminant is non-negative, i.e., $D=36(2 y-1)^{2}-12\left(14 y^{2}-20 y+11\right) \geq 0$. After transformation, we get: $y^{2}-4 y+4 \leq 0$. From which $\boldsymbol{y}=2$. Substituting the found value of $\boldsymbol{y}$, 2, into the equation, we obtain the quadratic equation: $x^{2}-6 x+9=0$. From which $\boldsymbol{x}=3$. Therefore, the solution to the equation is: $\boldsymbol{x}=3, \boldsymbol{y}=2$.