complete set of values of $a$ having modulus and linear terms 
If $(9-x^2)>|x-a|$ has at least one negative  real solution for $a\in\mathbb{R}.$ Then complete set of values of $a$ is 

Plan
If $x>a$ Then $9-x^2>x-a\Rightarrow x^2+x-(a+9)<0$
If $x\leq a$ Then $9-x^2>a-x\Rightarrow x^2-x+a-9<0$
How do i solve these inequalityHelp me please

For the case $x>a$:
$$x^{2} + x - (a+9) < 0 $$
The roots are:
$$\frac{-1 \pm \sqrt{4a+37}}{2} $$
Notice $a \ge -37/4$. Clearly the solution is:
$$\frac{-1 - \sqrt{4a+37}}{2} < x < \frac{-1 + \sqrt{4a+37}}{2} $$
The least possible value for $x$ is therefore $-1/2$, and clearly $-1/2 > -37/4$, so $a \ge -37/4$ does not violates $x>a$. So, this solution has at least one negative value, $x=-1/2$, that is when $a=-37/4$. Notice that for this case it is always $\frac{-1 - \sqrt{4a+37}}{2} < 0$. So all feasible values of $a$ is $a \ge \frac{-37}{4}$.

For the case $x \le a$, the method is similar.
