Maximum value of $|x-y|$ Given $x,y\in\mathbb R$ such that
$$5x^2+5y^2-6xy\ =\ 8$$
find the maximum value of $|x-y|$.
My attempt
$5x^2 - 6yx + (5y^2-8)\ =\ 0$
$x\ =\ \dfrac{6y\pm\sqrt{(6y)^2-4(5)(5y^2-8)}}{10} = \dfrac{6y\pm\sqrt{160-64y^2}}{10} = \dfrac{3y\pm2\sqrt{10-4y^2}}{10}$
$5y^2 - 6xy + (5x^2-8)\ =\ 0$
$y\ =\ \dfrac{3x\pm2\sqrt{10-4x^2}}{10}$
$|x-y|\ =\ \dfrac{\left|3(y-x)\pm2\sqrt{10-4y^2}\mp2\sqrt{10-4x^2}\right|}{10}$
What do I do from here?

You tried to solve for both $x$ in terms of $y$ and $y$ in terms of $x$.
Try expressing everything as either a function of $x$ or of $y$.
