Four. (30 points) As shown in the figure, $AC$ is a vertical pole standing on a flat ground, and $AB$ is a taut steel cable used to secure the pole. It is known that the length of the steel cable is $l$, and the point $D$ on the steel cable is at a distance $d$ from both the pole and the ground. Try to express the distances from the two points where the cable touches the ground to the foot of the pole, $h_{1}, h_{2}$, in terms of $d$ and $l$.

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Four, $\because \triangle A D E \sim \triangle A B C$,
$$
\therefore \frac{D E}{B C}=\frac{A E}{A C}, \frac{d}{h_{2}}=\frac{h_{1}-d}{h_{1}},
$$

It is known that $h_{1}+h_{2}=\frac{h_{1} h_{2}}{d}$. Let the value of the above expression be $m$. By Vieta's formulas, $h_{1}, h_{2}$ are the roots of the equation $x^{2}-m x+d m=0$, thus we have
$$
x=\frac{1}{2}\left(m \pm \sqrt{m^{2}-4 d m}\right).
$$

Squaring and rearranging (1) gives $h_{1}^{2}+h_{2}^{2}=\frac{\left(h_{1} h_{2}\right)^{2}}{d^{2}}-2 h_{1} h_{2}$, which is
$$
l^{2}=m^{2}-2 d m,
$$

Thus, $m^{2}-2 d m-l^{2}=0$,
$$
\begin{aligned}
\therefore m & =\frac{2 d \pm \sqrt{4 d^{2}+4 l^{2}}}{2} \\
& =d \pm \sqrt{d^{2}+l^{2}},
\end{aligned}
$$

Discarding the negative value, substituting the positive value into (2), and simplifying, we get
$$
\begin{aligned}
x= & \frac{1}{2}\left(d+\sqrt{d^{2}+l^{2}}\right. \\
& \pm \sqrt{\left.l^{2}-2 d^{2}-2 d \sqrt{d^{2}+l^{2}}\right)}
\end{aligned}
$$

as the values of $h_{1}, h_{2}$.