Let $u = x - \frac32.$ Then
&= \left(u^2 - \frac94\right)\left(u^2 - \frac14\right).
Let $v = u^2.$ Find the minimum value of the quadratic polynomial
$\left(v - \frac94\right)\left(v - \frac14\right)$ on the domain $v \geq 0.$
(It occurs when $v = \frac12\left(\frac94 + \frac14\right).$)
Working backwards, this is also the minimum value of $x(x−1)(x−2)(x−3).$
The fact that the minimum of $x(x−1)(x−2)(x−3)$ occurs at two different points means that if you translate the graph up just far enough so that it touches the $x$-axis at its two minima, you will have two double roots, which means all your roots are real (there are only four roots). If you translate the graph any higher then it doesn't meet the $x$ axis and you have no real roots at all.
As has already been pointed out, there are other translations of the graph that put its minima and a local maximum below the $x$-axis, leaving only two $x$-intercepts which give you two real roots (but not double roots, so the other two must be complex).
The symmetry of the polynomial
says there should be a local minimum or maximum at $u=0$;
in fact, it's a local maximum.