Let $Q(x) = x^3 - 12x$. Then $Q'(x) = 3x^2 - 12$ has roots $-2$ and $2$, and $Q$ has a local minimum at $(2, -16)$ and a local maximum at $(-2, 16)$. So $Q(x) = -t$ has three (not necessarily distinct) real roots for $-16 \le t \le 16$ and one real root for $t < -16$ and $t > 16$, which means that $\Delta(t) = 0$ for $t < -16$ or $t > 16$. So from now on we consider only $t \in [-16, 16]$.

Let $u \le v \le w$ be the roots. Since $P'_t(x) = 0 \iff x = -2$ or $x = 2$ we have $u \le -2 \le v \le 2 \le w$. In particular, $-2 \le v \le 2$, and $v$ can assume any value in this interval: if $t = -16$, $v = -2$ and if $t = 16$, $v = 2$.

But we know that $u + v + w = 0$ and $uv + vw + uw = -12$, so $u + w = -v$ and $uw = -12 - v(u + w) = v^2 - 12$, so $(\Delta(t))^2 = (w - u)^2 = (w + u)^2 - 4uw = 48 - 3v^2$, which lies in the range $[48 - 3 \cdot 2^2, 48] = [36, 48]$. So $36 \le (\Delta(t))^2 \le 48 \iff 6 \le \Delta(t) \le 4\sqrt{3}$.

So the range of $\Delta(t)$ is $\{0\} \cup [6, 4\sqrt{3}]$.