<s>[INST] Factorization Manipulation Let $x$ and $y$ be real numbers. Consider $t=x^2+10y^2-6xy-4y+13$. So what is the $t$ as smallest number? The solution is $9$.
My trying: 
$$t=x^2+10y^2-6xy-4y+13$$
$$=x^2+9+10y^2-6xy-4y+4$$
   $$=(x+3)^2-3x+2y(5y-2)-6xy+4$$.
So let $x=-3$ and $y=0$. From this I got $t=13$. My answer is false. Can you help?
 [/INST] Notice that
$$t=x^2+10y^2-6xy-4y+13=x^2-6y \cdot x+10y^2-4y+13,$$which could be regarded as a quadratic function with respect to the variable $x$.
Hence, $$t \geq \dfrac{4\cdot 1 \cdot (10y^2-4y+13)-(-6y)^2}{4 \cdot 1}=y^2-4y+13=(y-2)^2+9\geq 9,$$ with the equality holding if and only if $x=-\dfrac{-6y}{2 \cdot 1}$ and $y=2$,namely, $x=6,y=2$.
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