8. (4 points) The decimal representation of a four-digit number ABCD, which does not contain nines, was divided into three parts in three ways, after which three quadratic equations were formed: $A x^{2} + B x + B \Gamma = 0$, $A x^{2} + B B x + \Gamma = 0$, and $AB x^{2} + B x + \Gamma = 0$. It turned out that all of them have roots. Find all possible values of ABCD.

AB, BC, and CD here are not products of the digits, but two-digit numbers. Not only in the representation of the number ABCD, but also in the representations of these two-digit numbers, the first digits cannot be equal to 0.
Answer: 1710 or 1810.

Solution: From the first equation, we get that $B^2 \geq 4 \cdot A \cdot B\Gamma$. Since according to the condition $B^2 \leq 64$, we get that $B\Gamma \leq 16$, which means $B=1$ (since zero cannot be equal to any digit except $\Gamma$).

On the other hand, from the third equation, we get $1=B^2 > 4 \cdot AB \cdot \Gamma$, which is only possible when the right side is 0, that is, when $\Gamma=0$. This condition ensures that the second and third equations have roots.

Returning to the first equation: $64 \geq B^2 \geq 40 \cdot A$. From this, $A=1$, and B is 7 or 8.