Let $f = 2x^4 + 2(a - 1)x^3 + (a^2 + 3)x^2 + bx + c.$ ,Find out $a, b, c ∈ R$ and its roots knowing that all roots are real. 

Let $f = 2x^4 + 2(a - 1)x^3 + (a^2 + 3)x^2 + bx + c.$
Find out $a, b, c ∈ R$  and its roots knowing that all roots are real.



The first thing that came into my mind was to use vieta's formulas so let roots be $\alpha , \beta , \gamma , \delta$
$$ \alpha + \beta +\gamma+\delta=-\frac{b}{a}$$
$$ \Rightarrow \alpha + \beta +\gamma+\delta = 1-a$$
$$ \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma+\beta\delta+\gamma\delta=\frac{c}{a}$$
$$ \Rightarrow \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma+\beta\delta+\gamma\delta =\frac{a^2+3}{2}$$
$$ \alpha\beta\gamma + \alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=\frac{-d}{a}$$
$$ \Rightarrow  \alpha\beta\gamma + \alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta= \frac{-b}{2}$$
$$ \alpha\beta\delta\gamma = \frac{e}{a} $$
$$ \Rightarrow \alpha\beta\delta\gamma = \frac{c}{2} $$
But that did not get me anywhere...
Also I took the second derivative and set it equal to zero but that led me to complex roots... any ideas?

Hint: Since $f(x)$ has $4$ roots, the equation $f''(x)=0$ has at least one solution. Note that 
$$
f''(x)=24x^2+12(a-1)x+2(a^2+3)
$$
The discriminate is 
$$
\Delta=12^2(a-1)^2-4\cdot 24\cdot 2(a^2+3)=-48(a+3)^2
$$
In order to have a root we must have that $a=-3$. 
