4. From $1,2, \cdots, 10$ choose 3 different numbers $a, b, c$ as the coefficients of the quadratic equation $a x^{2}+b x=c$. Then the number of equations with different solutions is $\qquad$
4.654.

When considering $a1$, for each factor $k$ of $d$ greater than 1, if $\frac{c}{k} \geqslant 3$, then the array $\left(\frac{a}{k}, \frac{b}{k}, \frac{c}{k}\right)$ has already been considered before and should be removed. The number of remaining arrays is denoted as $f(c)$. Thus, we have
$$
\begin{array}{l}
f(3)=1, f(4)=3, f(5)=6, f(6)=10-f(3)=9, \\
f(7)=15, f(8)=21-f(4)=18, \\
f(9)=28-f(3)=27, f(10)=36-f(5)=30 .
\end{array}
$$

Therefore, $f(3)+f(4)+\cdots+f(10)=109$.
For each group $(a, b, c)$, by swapping the positions of $a$, $b$, and $c$, 6 different solutions of the equation can be formed.
Thus, the total number of equations is $109 \times 6=654$.