I offer a long, tedious, boring solution. I hope there is an elegant proof of this problem, and someone will post it.
Suppose that the roots of
$$p(z)=2\,\left(z^4-\frac{3}{2}\,z^3+\frac{3}{2}\,z^2-\frac12\,z+\frac12\right)$$
are $a\pm b\text{i}$ and $c\pm d\text{i}$ for some $a,b,c,d\in\mathbb{R}$ (we already know that $p(z)$ has no real roots). We may further assume that $b$ and $d$ are positive (but this is irrelevant to the proof of this problem). Thus,
$$p(z)=2\,\big((z-a)^2+b^2\big)\,\big((z-c)^2+d^2\big)\,,$$
so that
$$\begin{align}p(z)&=2\,\Biggl(z^4-2(a+c)\,z^3+\left(a^2+b^2+c^2+d^2+4ac\right)\,z^2\\&\phantom{aaaaa}-2\Big(\left(a^2+b^2\right)c+\left(c^2+d^2\right)a\Big)\,z+\left(a^2+b^2\right)\left(c^2+d^2\right)\Biggr)\,.\end{align}$$
Consequently, if $m:=ac$, $x:=a^2+b^2$, and $y:=c^2+d^2$, then
$$a+c=\frac34\,,\tag{1}$$
$$x+y=\frac{3}{2}-4m\,,\tag{2}$$
$$cx+ay=\frac14\,,\tag{3}$$
and
$$xy=\frac12\,.\tag{4}$$
From (2) and (3), we get
$$(a-c)x=\frac{3}{2}a-4ma-\frac14\text{ and }(c-a)y=\frac{3}{2}c-4mc-\frac14\,.$$
Multiplying the two results above and using (4), we obtain
$$-\frac{1}{2}(a-c)^2=\left(\frac{3}{2}a-4ma-\frac14\right)\left(\frac{3}{2}c-4mc-\frac14\right)\,.$$
From (1), we can see that the previous equation is equivalent to
$$-\frac{1}{2}\left(\left(\frac{3}{4}\right)^2-4m\right)=16m^3-12m^2+3m-\frac{7}{32}\,.$$
Hence, $m$ is a real root of the cubic polynomial
$$q(t):=16t^3-12t^2+t+\frac{1}{16}\,.$$
Note hat $q'(t)=48t^2-24t+1$, so $q'(t)$ has two roots $$\dfrac{3-\sqrt{6}}{12}>0.045>\dfrac{1}{40}\text{ and }\dfrac{3+\sqrt{6}}{12}<0.455\,.$$ Since $\lim\limits_{t\to-\infty}\,q(t)=-\infty$, $q(0)=\dfrac{1}{16}>0$, $q\left(\dfrac12\right)=-\dfrac{7}{16}$, and $\lim\limits_{t\to+\infty}\,q(t)=+\infty$, we conclude that $q(t)$ has three real roots $u<0$, $v\in (0.045,0.455)$, and $w>\dfrac12$. (Numerically, $u\approx -0.04111$, $v\approx 0.14768$, and $w\approx 0.64343$.)
We may assume without loss of generality that $a\geq c$. Since $a+c=\dfrac34$, it follows immediately that $a>0$. Since $0$ is not a root of $q(t)$, $c\neq 0$. We shall prove that $c<0$ and the claim that $p(z)$ has one complex root in each of the four quadrants is established.
Suppose for the sake of contradiction that $c>0$, then by the AM-GM Inequality, we have
$$\frac{1}{2}=xy\leq \left(\frac{x+y}{2}\right)^2=\left(\frac{\frac32-4m}{2}\right)^2\text{ and }m\leq \left(\frac{a+c}{2}\right)^2=\frac{9}{64}\,.$$
That is, $m\leq \dfrac{9}{64}$, and either
$$m\geq \frac{3}{8}+\frac1{2\sqrt{2}}\text{ or }m\leq \frac{3}{8}-\frac1{2\sqrt{2}}\,.$$
Therefore,
$$m\leq \frac{3}{8}-\frac{1}{2\sqrt{2}}<0.022<\frac{1}{40}\,.$$
However, since $m$ is a root of $q(t)$ and $m>0$, it must equal $v$ or $w$, but both $v$ and $w$ are greater than $\dfrac1{40}$. This is a contradiction, and so $m=u<0$ must hold. This means $c<0$. WolframAlpha confirms this: $a\approx 0.80130$, $b\approx 0.79308$, $c\approx -0.05130$, and $d\approx 0.62509$.