\(A=x^2+y^2-x+6y+10=x^2-x+\frac{1}{4}+y^2+6y+9+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\)
\(\left(x-\frac{1}{2}\right)^2\ge0;\left(y+3\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(MinA=\frac{3}{4}\Leftrightarrow x=\frac{1}{2};y=-3\)