What is the largest $a$ for which all the solutions to the equation $3x^2+ax-(a^2-1)=0$ are positive and real? Problem: What is the largest $a$ for which all the solutions to the equation $3x^2+ax-(a^2-1)=0$ are positive and real?
Attempt: Solving the equation for $x$ I get $$x_{1,2}=-\frac{a}{6}\pm\sqrt{\frac{13a^2-12}{36}}.$$
For the roots to be real, the discriminant as to be greater than or equal to zero, so it yields the inequality 
$$13a^2-12 \geq 0\Leftrightarrow -\frac{2\sqrt{39}}{13}\leq a\leq\frac{2\sqrt{39}}{13}.$$
Condition number two is that both roots should be positive. How should I think to proceed?

First of all, the inequality you end up with is wrong: you need either
$$a \leq -\frac{2\sqrt{39}}{13} \;\;\;\text{     or     }\;\;\; a \geq \frac{2\sqrt{39}}{13}$$
Since the root 
$$x_2 = -\frac{a}{6} - \sqrt{\frac{13a^2 - 12}{36}} = \frac{1}{6}\left(-a - \sqrt{13a^2 - 12}\right)$$
is always smaller than $x_1$ (when both are real, of course), it suffices to find the largest $a$ such that $x_2$ is positive. 
Hence we are looking for the largest $a$ such that
$$-a - \sqrt{13a^2 - 12} \geq 0.$$
Now, as the root is always positive and minus the root is therefore negative, obviously we'll need $a \leq 0$. Rearrange as
$$-a \geq \sqrt{13a^2 - 12}$$
and square both sides to find
$$a^2  \geq 13a^2 - 12,$$
and rearrange again to find 
$$12 \geq 12a^2$$
or hence
$$1\geq a^2.$$
Since $a$ must be negative, this means that $-1 \leq a \leq 0$.
Therefore, the values of $a$ such that the equation has $2$ (not necessarily distinct) real roots which are both positive, are
$$a \in \left[-1,-\frac{2\sqrt{39}}{13}\right]$$
hence the largest such $a$ is $-\frac{2\sqrt{39}}{13}$
