\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|} \hline 110 & $12 \square$ & $13 \square$ & $14 \square$ & $15 \square$ & $16 \square$ & $\mathbf{1 7}$ & $\mathbf{1 8}$ & $\mathbf{1 9}$ & $\mathbf{2 0}$ \\ \hline \end{tabular} answer. a. $\left(-\infty,-\frac{57}{16}\right]$ b. $\left[-\frac{57}{16}, \infty\right)$ c. $\left(-\infty,-\frac{5}{8}\right]$ d. $\left[-\frac{5}{8}, \infty\right)$

Step 1 ：Simplify the expression: \(12x^2 + (13 + 14 + 15 + 16)x - (18 + 19 + 20)\)

Step 2 ：Calculate the solutions: \(x = -\frac{29}{12} + \frac{5\sqrt{61}}{12}\) and \(x = -\frac{5\sqrt{61}}{12} - \frac{29}{12}\)

Step 3 ：Determine the intervals: \(x = -\frac{29}{12} + \frac{5\sqrt{61}}{12}\) belongs to option d, and \(x = -\frac{5\sqrt{61}}{12} - \frac{29}{12}\) belongs to option a

Step 4 ：Final Answer: \(\boxed{a. \left(-\infty,-\frac{57}{16}\right]}\) and \(\boxed{d. \left[-\frac{5}{8}, \infty\right)}\)