Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$ in real numbers 
Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$

The problem does not say it but I think solutions should be from $\mathbb{R}$. I tried to express the left sum as a sum of squares but that does not work out. Any suggestions?

You can solve for $x$:
$(2)x^2+(6y-2)x+(5y^2-4y+1)=0\implies$
$x_{1,2}=\frac{-(6y-2)\pm\sqrt{(6y-2)^2-4\cdot2\cdot(5y^2-4y+1)}}{2\cdot2}=\frac{-6y+2\pm\sqrt{-4y^2+8y-4}}{4}=\frac{-6y+2\pm\sqrt{-4(y-1)^2}}{4}=\frac{-6y+2\pm2i(y-1)}{4}$
Then the only real solution is with $y=1$, hence $x=-1$.
