2. The equation
$$
\sqrt[3]{(x+7)(x+8)}-\sqrt[3]{(x+5)(x+10)}=2
$$

has $\qquad$ real solutions $x$ that are not equal.
2.4.

Let $a=\sqrt[3]{(x+7)(x+8)}$,
$$
b=\sqrt[3]{(x+5)(x+10)} \text {. }
$$

Then $a-b=2, a^{3}-b^{3}=6$.
Eliminating $a$ gives
$$
\begin{array}{l}
3 b^{2}+6 b+1=0 \\
\Rightarrow b_{1}=\frac{-3+\sqrt{6}}{3}, b_{2}=\frac{-3-\sqrt{6}}{3} .
\end{array}
$$

If $b_{1}=\frac{-3+\sqrt{6}}{3}$, then
$$
x^{2}+15 x+50+\left(\frac{3-\sqrt{6}}{3}\right)^{3}=0 \text {. }
$$

By $\Delta_{1}=15^{2}-4\left(50+\left(\frac{3-\sqrt{6}}{3}\right)^{3}\right)>0$, we know the original
equation has two distinct real roots $x_{1} 、 x_{2}$.
If $b_{2}=\frac{-3-\sqrt{6}}{3}$, then
$$
x^{2}+15 x+50+\left(\frac{3+\sqrt{6}}{3}\right)^{3}=0 \text {. }
$$

By $\Delta_{2}=15^{2}-4\left(50+\left(\frac{3+\sqrt{6}}{3}\right)^{3}\right)>0$, we know the original equation has two distinct real roots $x_{3} 、 x_{4}$.

Since $x_{1} 、 x_{2} 、 x_{3} 、 x_{4}$ are all distinct, the number of distinct real solutions $x$ of the original equation is 4.