Find all real numbers m such that $x^2+m{y}^2-4my+6y-6x+2m+8\ge 0$ If K is the largest value of m then find the value of 2K is______.

This looks rather complicated. one nice thing about this inequality is that the highest degree terms of x and y are both of degree 2, which allows us to use properties of quadratic functions. Interpret the inequality as a quadratic function in x. Rearranging terms, we have

$x^2-6x+m{y}^2-4my+6y+2m+8\ge 0.$
Since the left hand side is nonnegative, and the leading coefficient is positive, it follows that the discriminant, Δ, of this quadratic in x must be nonpositive. In particular, for all y, we must have
$\mathrm{\Delta }=6^2-4(m{y}^2-4my+6y+2m+8)\le 0.$

This is still a fairly complicated condition. But again, the highest degree term of y is of degree 2, so we can follow the same strategy and interpret the discriminant as a quadratic in y. Since the quadratic must be nonpositive for all y, the leading term must be negative, which implies $m>0.$. As a quadratic in y, we can rearrange terms to obtain
$\mathrm{\Delta }=4(-m{y}^2+2y(2m-3)-2m+1)\le 0.$
We can repeat the process, and take the discriminant of our nonpositive quadratic in y. The new discriminant function, Δ^', should be nonpositive. Hence
${\mathrm{\Delta }}^{\mathrm{\prime }}=4\left(\right(2m-3{)}^2-m(2m-1))\le 0$

This is equivalent to a quadratic inequality in m, that is
$2m^2-11m+9=2(m-1)(m-\frac{9}{2})\le 0$