$3x^2-12x+m$ is negative for all values of $x$ The question says that find the range of value of $m$ for which the function $3x^2-12x+m$ is negative for all values of $x$
My attempt, 
$b^2-4ac<0$
$(-12)^2-4(3)(m)<0$
$144-12m<0$
$m>12$
But the given answer is $m<12$. Why?
 A: From this graph, it is clear that by having $m>12$, we will have no real solutions, and no negative values for the function at any point $x$.
I think the question might be poorly written, and it could be asking, for which values of $m$, can the function attain negative values. This would give you $m < 12$.
A: $b^2-4ac<0$ means that $ax^2+bx+c=0$ has no real solution, that is, $ax^2+bx+c\ne 0, \forall x.$ Since your parabola opens up, you have that $3x^2-12x+m$ is always positive under your assumption.
So, if you are interested into $ax^2+bx+c$ being negative for some $x,$ you need $b^2-4ac>0,$ since your parabola opens up. In such a case, you get $m<12.$
A: If you rewrite your proof for $b^2-4ac>0$, instead of for $b^2-4ac<0$, you will obtain the desired result.  
What you have discovered is values for $m$, that provide real roots.
Since the parabola is curved upwards, it will always be positive for some value of $x$. However, by finding the range of $m$ that produces a function with real roots, you've found the forms of this parabola that dip below the $x$-axis. You've found values of $m$ that produces some negative values for y.  
(I question the wording of the question here. Is this exactly how it's posed in the textbook, or wherever this question came from?)  
You can play with this function here. Try moving the $m$ slider and see what happens to the parabola. 
Here's a snapshot where $m=12$.  

