[tex]4x^2+9y^2+12(x-y)+13 \leq 0\\
4x^2+9y^2+12x-12y+13\leq 0\\
(4x^2+12x+9)+(9y^2-12y+4)\leq 0\\
(2x+3)^2+(3y-2)^2\leq 0\\
\text{Deoarece }(2x+3)^2\geq 0\ \text{si}\ (3y-2)^2\geq 0~\text{inseamna ca:}\\ (2x+3)^2+(3y-2)^2\geq 0,\ \text{ceea ce implica faptul ca:} \\
\left \{ {{2x+3=0} \atop {3y-2=0}} \right.\Leftrightarrow x=-\dfrac{3}{2}; y=\dfrac{2}{3}[/tex]
