Find all real values that the parameter $m$ can take, knowing that the solutions of the equation

$$
x(x-1)(x-2)(x-3)=m
$$

are all real.
$\left[-1, \frac{9}{16}\right]$

We trace the curve of the equation $y=x(x-1)(x-2)(x-3)$. The parameters $m$ that work are those for which the curve intersects the line of the equation $y=m$ at 4 points (in case of tangency, one point counts double). It is sufficient to determine the local extrema; beyond these, there will not be enough points of intersection, and thus there will be non-real complex solutions to our equation. Here they are:

$-\frac{9}{16}:$ reached at $\frac{3}{2}$

--1 : reached at $\frac{3-\sqrt{5}}{2}$

--1 : reached at $\frac{3+\sqrt{5}}{2}$