15. Given $A, B, C, D \in \{1,2, \cdots, 6\}$, and they are all distinct. If the curves $y=A x^{2}+B$ and $y=C x^{2}+D$ intersect, then the number of different ways to choose $A, B, C, D$ is $(\quad)$ (the intersection of the curves is independent of the order of $A, B, C, D$, for example, $A=3$, $B=2, C=4, D=1$ and $A=4, B=1, C=3, D=2$ are considered the same selection).
(A) 30
(B) 60
(C) 90
(D) 180
(E) 360
15. C.

From the problem, we know
$$
A x^{2}+B=C x^{2}+D \Rightarrow(A-C) x^{2}=D-B
$$

has real roots.
Since \( A, B, C, D \) are all different, we have
$$
x^{2}=\frac{D-B}{A-C}.
$$

Thus, \( A-C \) and \( D-B \) must have the same sign.
Choosing two sets of numbers from \(\{1,2, \cdots, 6\}\), each set containing two numbers, there are \(\frac{1}{2} \mathrm{C}_{6}^{2} \mathrm{C}_{4}^{2}\) ways to choose; placing them in the denominator or numerator position has 2 ways.
Therefore, the required number is \(\frac{1}{2} \mathrm{C}_{6}^{2} \mathrm{C}_{4}^{2} \times 2=90\).