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?

$$2x^2 + 5y^2 + 6xy -2x -4y+1=0$$
$$(1+1)x^2 + (4+1)y^2 + (4+2)xy - 2x -4y + 1=0$$
$$(x^2 +4y^2 +4xy -2x -4y + 1) + (x^2 +2xy + y^2)=0$$
$$(x+2y-1)^2 + (x+y)^2=0$$
Your idea to write the expression as a sum of squares is good intuition. Now, when can a sum of squares be zero? Exactly when both of the squares are zero.
Thus
$$x+2y-1=0$$
$$x+y=0$$
You should be able to find the solution from here.
