roots of a fourth degree polynomial function (Vieta) Question
I am interested in the root of the polynomial function :
$x^4+(a+b+c+d-2)x^3+(ab+ac+ad+bc+bd+cd-2b-2c-a-d)x^2
+(abc+abd+acd+bcd-ab-ac-ad-2bc-bd-cd-a-d+b+c)x+
abcd-abc-bcd-ad+bc=0$
Under the restriction that: $0<a,b,c,d<1$.
I tried Vietas formula but don’t know how to come to a decomposition
$a_0=abcd-abc-bcd-ad+bc=x_1x_2x_3x_4$

There are two real solutions, the general expressions are very, very long. Even if all four coefficients are equal, $a=b=c=d$, you have an equation
$$a^4-2 a^3+(4 a-7) a^2 x+(4 a-2) x^3+6 (a-1) a x^2+x^4=0$$
with complicated solutions.

 For example, if all coefficients are equal to 1/2 the two real solutions are
$$x=\frac{4+6^{2/3}\pm\sqrt{20 \sqrt{4+6^{2/3}}-6 \sqrt[3]{6}+4\ 6^{2/3}+32}}{4 \sqrt{4+6^{2/3}}}.$$
