Sửa đề: \(2x^2+y^2+6x-4y=\frac{-17}{2}\)
Ta có: \(2x^2+y^2+6x-4y=\frac{-17}{2}\)
=>\(2x^2+6x+\frac92+y^2-4y+4=0\)
=>\(2\left(x^2+3x+\frac94\right)+\left(y-2\right)^2=0\)
=>\(2\left(x+\frac32\right)^2+\left(y-2\right)^2=0\)
=>\(\begin{cases}x+\frac32=0\\ y-2=0\end{cases}\Rightarrow\begin{cases}x=-\frac32\\ y=2\end{cases}\)