Real values of $a$ for $x^4+(a-1)x^3+x^2+(a-1)x+1=0$ to have at least two negative roots 
Problem Statement:-
Find all the values of $a$ for which the equation 
  $$x^4+(a-1)x^3+x^2+(a-1)x+1=0$$
  possess at least two negative roots.


I know that there has been a post regarding this same problem here, but I have a different issue with the problem than that posted in the OP in the above post. 
So, I don't have an issue of finding a solution and coincidentally I had the same approach as the one posted by RicardoCruz. 
But still if you have a solution than you are more than welcome to post one, well come on who doesn't want some good reputation change ;P.

My Attempt at a solution:-

For those who wanna skip to my main issue may start to read from the next quote.

We can see very clearly that $x=0$ isn't a root of the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0$$
hence it is safe to divide the equation throughout by $x^2$ for arriving at the condition which the problem wants.
So, we get
$$\left(x^2+\dfrac{1}{x^2}+2\right)+(a-1)(x+\dfrac{1}{x})-1=0\\
\implies \left(x+\dfrac{1}{x}\right)^2+(a-1)(x+\dfrac{1}{x})-1=0$$.
Now, let $z=x+\dfrac{1}{x}$ then the equation becomes
$$z^2+(a-1)z-1=0\\
\implies z=\dfrac{-(a-1)\pm\sqrt{\left(a-1\right)^2+4}}{2}$$
Now, its necessary to note that $\sqrt{\left(a-1\right)^2+4}\gt(a-1)$, hence for the roots to be negative $$\left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}$$
As, we know that $\text{A.M. $\ge$ G.M.}$ for non negative real numbers, and as we are dealing with negative roots of the equation, so we will be applying $\text{A.M. $\ge$ G.M.}$ to $-x,-\dfrac{1}{x}$, we get 
$$\dfrac{-\left(x+\dfrac{1}{x}\right)}{2}\ge1\implies x+\dfrac{1}{x}\le-2\\
\implies \left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\le -2$$
But as the roots are supposed to be distinct hence, we have to exclude the case of equality, because it occurs when $x=\dfrac{1}{x}=-1$, which results in $x+\dfrac{1}{x}=-2\implies (x-1)^2=0$, hence $x$ has a repeated root. So, we get
$$\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\lt -2\implies \sqrt{\left(a-1\right)^2+4}\gt 5-a$$

Now, this is where I am having problem, that's right solving the "inequality". I always had thought of this before but never tried to test it before, so bear with it and follow me on to a trip to solve inequalities in a weird way.

I did a case study for the inequality, which is as follows:-
Case-1:-
If $(5-a)\gt 0\implies a\lt 5$, then there is no problem in squaring the inequality as both sides are positive.
$$\left(a-1\right)^2+4\gt a^2+25-10a\implies a\gt \dfrac{5}{2}$$
$$\therefore a\in\left(\dfrac{5}{2},5\right)$$
Case-2:-
If $5-a\lt0$, then we might be faced with a problem cause, there might be a case where $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$, which would make the inequality $\sqrt{\left(a-1\right)^2+4}\gt 5-a$ on squaring change the sign of inequality to be in accordance with the condition $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$.
So, on solving the inequality we get,
$$\sqrt{\left(a-1\right)^2+4}\gt 5-a\implies (a-1)^2+4\lt(5-a)^2$$
Now, why did I change the sign of inequality was due to the fact that I was finding the interval where(as per the condition I stated before) $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$. So, just like $5\gt-6$ but on squaring the inequality we get $25\lt36$. So that's what I did.
This results in $a\lt\dfrac{5}{2}\cup a\gt5\implies a\in\emptyset$
So, on combining both the cases we get $a\in\left(\dfrac{5}{2},5\right)$
But from RobertCruz's solution and the books answer it seems that $ a\in\left(\dfrac{5}{2},\infty\right)$ is the correct solution, so what am I doing wrong here.

Just posting an alternative solution, as the comments already helped you to debug your solution:
For quartic polynomials, there is a closed-form analytical solution for the zeroes. However horrible that formula may seem, you don't need it here. You only need the discriminant.
A polynomial $ax^4+bx^3+cx^2+dx+e=0$ has two real solutions, when the discriminant is negative:
$$\Delta=256 a^3 e^3 - 192 a^2 b d e^2 - 128 a^2 c^2 e^2 + 144 a^2 c d^2 e$$
$$ - 27 a^2 d^4 + 144 a b^2 c e^2 - 6 a b^2 d^2 e - 80 a b c^2 d e + 18 a b c d^3 $$
$$+ 16 a c^4 e - 4 a c^3 d^2 - 27 b^4 e^2 + 18 b^3 c d e - 4 b^3 d^3 - 4 b^2 c^3 e + b^2 c^2 d^2<0$$
In your case, the polynomial can be written also as
$$\underbrace{x^4+x^2+1}_{>0} + (a-1)\underbrace{(x+x^3)}_{>0 \text{ if } x>0}=0$$
for which any potential roots are negative if $a-1>0$ and positive otherwise.
Therefore, having checking for $a-1>0$ and negative discriminant ensures two negative real roots. You get a third degree polynomial in $u=(a-1)^2$.
$$144+8u-23u^2-4u^3<0$$
It turns out this factors:
$$(4u-9)(4+u)^2>0$$
$$|u|>9/4$$
$$a-1>3/2$$
Also, you can quickly prove that your polynomial cannot have more than two real roots. Either by checking other tests on the link above (checking the value for $P$ and $D$) or by writing
$$x=-\frac{1}{1-a}\frac{1+x^2+x^4}{1+x^2}=-\frac{1}{a-1}\left(1+\frac{x^4}{1+x^2}\right)$$
RHS has no inflection points, so a straight line can intersect it twice at most (definition of convexity).
