Solution of an inequaility. I try to solve following inequailty $-3t^4-4Bt^3-2B^2t^2+(6D-2BC)t+2BD-C^2 \leq 0$ where $B$, $C$ and $D$ are real numbers.
Say $g(t)=-3t^4-4Bt^3-2B^2t^2+(6D-2BC)t+2BD-C^2$. 
We observe that $g''(t)=-4(3t+B)^2$ is non-positive for all $t$. This means that $g(t)$ is concave down. Thus $g(t)$ has a local(absolute) maximum. If this absolute max. value is negative, the solution of this inequality is real numbers. Actually, I evaulated x-component of this point but still I didn't solve. 
 A: Your question is not really clear, but let's see if this helps.
Starting from your first derivative, we can find the critical points by setting:
$$-12t^3 - 12Bt^2 - 4B^2t - 6D - 2BC = 0$$
Diving by two:
$$-6t^3 - 6Bt^2 - 2B^2t - 3D - BC = 0$$
As a third degree equation, we can use Cardano's method neglecting the imaginary solutions, which in this case are two. The remaining solution is
$$t_0 =\frac{1}{3} \left(-\sqrt[3]{\frac{-2 B^3+9 B C+27 D}{2}}-B\right)$$
Considering that we have a cube root, we are not worried about the values of $B, C, D$.
At this point, we know, as you pointed out, from the second derivative that the function is negative $\forall t$, whence $t_0$ is itself a maximum point. In particular it's an absolute max point since it's unique.
To have the maximum value of the function, just substitute $t_0$ into $t$ in your initial function. A simple algebra leads thou to:
$$g(t_0) = \frac{1}{36} \left(-4 B^4-9 \left(5\ 2^{2/3} D \sqrt[3]{-2 B^3+9 B C+27 d}+4 C^2\right)-2\ 2^{2/3} B^3 \sqrt[3]{-2 B^3+9 B C+27 d}+9\ 2^{2/3} B C \sqrt[3]{-2 B^3+9 B C+27 D}+24 B^2 C\right)$$
Now, from here what do you really need? 
