Let's Get Back to A, B, C

From using the discriminant we obtain: $b^2 -4ac = b^2 +64a\leq\ 0$ . However, $b^{2}$ $\geq$ 0 , so ${a}$ will be negative and we want ${a}$ to have the largest value possible, so $b^2 -4ac = b^2 +64a = 0$ Now we rearrange this equation to obtain: $a = -\frac{b^2}{64}\Rightarrow\ 4a-b = -\frac{b^2}{16} - b = -(\frac{b^2 +16b}{16})$ This value is negtive and $b^{2}$ $\geq$ 0 so we must minimise ${b}$ (by completing the square) to maximise the whole expression. This gives: $4a-b = \frac{64 - (b+8)^2}{16}\rightarrow\ let \ b = -8 \rightarrow\ max \ (4a-b) = 4$

No distinct real roots means that the function is either always positive, or always negative, since it never cuts the x-axis.

We see that it is negative at $x=0$ , so it shall also be negative at $x=4$ . Plugging $x=4$ , we get $16a-4b-16<0$ , or $4a-b<4$ . Hence, the maximum possible value is $\boxed{4}$

its quite simple. no distinct real roots means F(x) is a square. but how, lets see $$ note that $(zix+yi)^2=-z^2x^2 -2zxy-y^2$ now inserting y=4 $(zix+4i)^2=-z^2x^2-8zx-16$ now we can say that $a=-z^2,b=8z$ now insert this in the expression $4a-b=-4z^2-8z$ now, the max will be attained at $z=-\dfrac{b}{2a}$ where $\quad a\quad$ is the coefficient of the 2nd degree term,and $b$ is the coefficient of the 1st degree term. $-\dfrac{b}{2a}=-\dfrac{-8}{2\times-4}=-1$ now insert this to get $-4(-1)^2-8(-1)=-4+8=\boxed{4}$

I also used the same method !

b^2+64a<=0. (discriminant <=0) 4a<=-(b^2)/16 4a max =-(b^2)/16 4a-b = -(b^2)/16-b =-(b^2+16b)/16 =-{(b+8)^2-64}/16 Max = 4 at b= -8

I think Lagrange Multipliers can also be used for this problem. :)