If $f(x)=4x^2+ax+a-3$ is negative for at least one negative $x$ find all possible values of $a$ 
If $f(x)=4x^2+ax+a-3$ is negative for at least one negative $x$ find all possible values of $a$

I don't know how to find all possible values.
I tried making the lower of the two roots as negative. That does not seem to work out. As $4>0$ the curve will be concave up. There can be numerous cases possible.

Examining the extrema of $f(x)=4x^2+ax+a-3$ which is a family of upwards open parabolas:
$$
f'(x) = 8 x + a = 0 \Rightarrow x = -\frac{1}{8}a \\
f''(x) = 8 > 0
$$ 
So we have a local minimum at $x = -(1/8)a$. There the function has the value:
$$
f(-(1/8)a) 
= \frac{1}{16}a^2 - \frac{1}{8}a^2 + a - 3 
= -\frac{1}{16}a^2 + a - 3
$$
The minima regarding different $a$ values lie on the curve
$$
m(x) = -4x^2-8x -3
$$

(Large image version)
From
$$
0 = -\frac{1}{16} a^2 + a - 3 \Rightarrow \\
a^2 = 16 a - 48 \Rightarrow \\
(a - 8)^2 = 64 - 48 = 16 \Rightarrow \\
a = 8 \pm 4
$$
we get that the minimum is non-negative for $a \in [4, 12]$ and there we will find no negative value for $f$ at all.
So we have a chance to find negative $f$ values for negative argument only for $a \in I = \mathbb{R} \setminus [4,12]$, where the minimum occurs for a negative function value. 
For positive $a\in I$ the minimum is at $x_m = -(1/8) a < 0$, so those $a$ are valid.
For negative $a \in I$ for $x = -1/2 < 0$ we have 
$f(-1/2) = 1 - a/2 + a - 3 = a/2 - 2 < 0$, so those $a$ are valid as well. 
This gives $I = \mathbb{R} \setminus [4,12]$ as set of all feasible $a$.
