Solve n7a^2-12a-8<0n4a^2-1<0

$$n 7 a ^ { 2 } - 12 a - 8 < 0 n 4 a ^ { 2 } - 1 < 0$$

Solve for n_7

$$n_{7}<\frac{12a+7}{a^{2}}$$

Solve for a

$$\left\{\begin{matrix}a\neq -\frac{7}{6}\text{, }&n_{7}=-\frac{36}{7}\\a\in \mathrm{R}\text{, }&n_{7}<-\frac{36}{7}\\a\in \left(-\frac{\sqrt{7n_{7}+36}-6}{n_{7}},\frac{\sqrt{7n_{7}+36}+6}{n_{7}}\right)\text{, }&n_{7}>0\\a>-\frac{7}{12}\text{, }&n_{7}=0\\a\in \left(-\infty,\frac{\sqrt{7n_{7}+36}+6}{n_{7}}\right)\cup \left(-\frac{\sqrt{7n_{7}+36}-6}{n_{7}},\infty\right)\text{, }&n_{7}>-\frac{36}{7}\text{ and }n_{7}<0\end{matrix}\right.$$