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?

Use Gauß' method to write a quadratic form as a linear combination of squares of linear forms:
\begin{align}
t&=x^2+10y^2-6xy-4y+13 =(x-3y)^2 -9y^2+10y^2-4y+13 \\
&=(x-3y)^2+y^2-4y+13 =(x-3y)^2+(y-2)^2-4+13\\
&=(x-3y)^2+(y-2)^2+9.
\end{align}
