Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$ Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$
I used the quadratic equation to get $$x=\frac{y-1\pm\sqrt{-2y-3y^2-\frac{5}{3}}}{4}$$
But I don’t see how that helps, hints and solutions would be appreciated 
Taken from the 2006 IWYMIC 

Hint: Show that $\frac{1}{3}$ is the unique minimum of the function $f(x,y)=(2x+1)^2+y^2+(y-2x)^2$. Then find the $(x,y)$ where this minimum occurs.
Your expression under the radical is also incorrect. It should be $-3y^2-2y-\frac{1}{3}$, or $-\frac{1}{3}(9y^2+6y+1) = -\frac{1}{3}(3y+1)^2$. Thus there is exactly one value of $y$ for which the expression under the radical is nonnegative.
