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
 A: Sketch a graph!
$k=5$">
Now you can see it more clearly. You want the red curve to be above the blue one.
For $a= 3$, the two curves intersect at $(3,0)$ and $(-2,5)$. (Solve the equation $9-x^2=-x+3$) So the solution set is $x\in (-2,3)$. The solution set is $x\in (-3,2)$ for $a=-3$.
For $a \leq 3$, we need to solve the equation $9-x^2=-x+a$:
$$x^2-x+a-9=0\\
B^2-4AC=1-4(a-9)=37-4a\\
B^2-4AC\geq 0\Rightarrow a\leq \frac{37}{4}.
x=\frac{1\pm\sqrt{37-4a}}{2}.$$
So the solution set is $x\in (\frac{1-\sqrt{37-4a}}{2},\frac{1+\sqrt{37-4a}}{2})$ for $3\leq a< \frac{37}{4}$, and $\emptyset$ for $a\geq \frac{37}{4}$.
By symmetry, the solution set is $x\in (\frac{-1-\sqrt{37+4a}}{2},\frac{-1+\sqrt{37+4a}}{2})$ for $-3\geq a> -\frac{37}{4}$, and $\emptyset$ for $a\leq \frac{37}{4}$.
It remains to consider $a\in(-3,3)$. You can work out that it is $(\frac{-1-\sqrt{37+4a}}{2},\frac{1+\sqrt{37-4a}}{2}).$
A: 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.
