<s>[INST] 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.
 [/INST] The function will be positive when $x$ is very large positive or very large negative. This means that there are two solutions, to your equation, of which at least one is negative.
The smallest root is:
$$\frac{-a -\sqrt{a^2-16(a-3)}}{8}$$
So we want to know when $a^2-16(a-3) > 0$, or $a^2-16a+48 > 0$, or $(a-8)^2-16 > 0$, or $(a-8)^2 > 16$, or $$a-8 > 4 \vee a-8 < -4$$
$$a > 12 \vee a < 4$$
When $a$ is positive, both $-a$ and $-a -\sqrt{a^2-16(a-3)}$ are negative, so then the smallest root is negative.
When $a$ is negative, $$\sqrt{a^2-16a+48}=\sqrt{(a-6)^2+12}>\sqrt{(a-6)^2}=|a-6|=6-a$$
With the the second because of the increasing nature of $\sqrt{x}$, the fourth because $a$ and thus $a-6$ negative is. Therefore we have $$\frac{-a -\sqrt{a^2-16(a-3)}}{8}<\frac{-a - (6-a)}{8}=\frac{-6}{8}=-0.75$$
Thus the solution is 
$$a > 12 \vee a < 4$$
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