I have a parametric quartic equation. It is potential of a black hole that I want to always be negative. I thought to make it to two quadratic equation. But it is very difficult to solve. What can I do?
My equation is: $-(\alpha) r^4 +(\alpha( L-a)^2 -2) r^2 +(L^2) r -2(L-a)^2$
This is black hole effective potential which I want to be always negative. α is the intensity of dark energy, a is spin of the black hole and L is particle angular momentum. For Kerr black hole effective potential is a quadratic equation and we know for always negative potential, discriminant must be negative when the coefficient of $r^2$ is negative. Then we get a relation for $L$. For this black hole, I want such $L$ in other variables except for $r$. It means $L(\alpha, a)$. ((In Kerr black hole it was $L(a)$.))
Prakhar's comment is essentially the answer. Whether or not it's helpful is up for debate.
Details: Given a quartic $ax^4+bx^3+cx^2+dx+e$, consider the following auxiliary variables:
\begin{align} \Delta\ =\ &256 a^3 e^3 - 192 a^2 b d e^2 - 128 a^2 c^2 e^2 + 144 a^2 c d^2 e - 27 a^2 d^4 \\ &+ 144 a b^2 c e^2 - 6 a b^2 d^2 e - 80 a b c^2 d e + 18 a b c d^3 + 16 a c^4 e \\ &- 4 a c^3 d^2 - 27 b^4 e^2 + 18 b^3 c d e - 4 b^3 d^3 - 4 b^2 c^3 e + b^2 c^2 d^2 \\ D =\ &8ac-3b^2 \\ P =\ &b^3+8da^2-4abc \end{align}
The quartic is negative everywhere if and only if: $\Delta$ is positive, and at least one of $P$ or $D$ is positive as well, and $a$ is negative. Notice that in your situation $b=0$, so we can simplify slightly:
\begin{align} \Delta\ =\ &256 a^3 e^3 - 128 a^2 c^2 e^2 + 144 a^2 c d^2 e - 27 a^2 d^4 \\ & + 16 a c^4 e \\ &- 4 a c^3 d^2 \\ D =\ &8ac \\ P =\ &8da^2 \end{align}
The simplification on $P$ and $D$ helps a bit: in particular we know that $P = 8\alpha^2L^2 \geq 0$, with equality if and only if $L=0$ or $\alpha=0$. I suspect that both of those conditions are unphysical, but if not, you'd have to do some thinking about $D$.
So $P$ is positive, and $a=-\alpha$ is negative, so we "only" need to check if $\Delta$ is positive. Unfortunately, substituting in the other variables gives a degree-10 inequality in $L$. It's possible that some careful and smart analysis can give you at least sufficient conditions that don't look horrifying. But especially without having physical intuition on this system, I'm inclined to look for a different path at this point.