A quadratic equation Find all values of a for which the quadratic $$\cos^2x - (a^2 + a + 5) |\cos x| + (a^3 + 3a^2 + 2a + 6) = 0$$ has real solution(s)
A. $a=-3$, B. $a=-2$, C. $a=-1$, D. $a=0$
I have solved this question by putting the values of the given options : 
Let $\cos x = t$
then it will look like $$t^2 - (a^2 + a + 5) |t| + (a^3 + 3a^2 + 2a + 6)$$
Comparing this quadratic with the standard quadratic and then finding out the discriminant for all four values of $a$ we get $D>0$ for all four values while the answer is given only $A$ and $B$
Kindly help.
 A: You have to check if for at least one root $t$ of the equation 
$$t^2 - (a^2 + a + 5) t + (a^3 + 3a^2 + 2a + 6)=
t^2 - (a(a+1)+ 5) t + (a(a+1)(a+2) + 6)
=0$$
there is a real $x$ such that $|\cos(x)|=t$.
Since $|\cos(x)|=t$ is solvable as soon as $t\in [0,1]$, just see if there is a root $t\in [0,1]$.
i) If $a=-3$ then $t^2 - 11t=0$ is solved by $t=0$ and $t=11$.
$0\in[0,1]$ so the answer is YES.
ii) If $a=-2$ then $t^2 - 7t+6=0$ is solved by $t=6$ and $t=1$.
$1\in[0,1]$ so the answer is YES.
iii) and iv) If $a=-1$ or $a=0$ then $t^2 - 5t+6=0$ is solved by $t=2$ and $t=3$.
No roots in in $[0,1]$ so the answer is NO.
A: Doing the same as Robert Z in his answer, considering  $$t^2 - (a^2 + a + 5) t + (a^3 + 3a^2 + 2a + 6)=0$$ the roots of the quadratic are $$t_1=3+a\qquad , \qquad t_2=2+a^2$$ since, after simplification $$\Delta=\left(a^2-a-1\right)^2$$ Obviousle $t_2$ must be discarded and $t_1$ would be acceptable if $3+a \leq 1$. So, if $a$ is suppose to be an integer, only $a=-2$ and $a=-3$ are acceptable.
