2. (25 points) Given the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two points $M\left(x_{1}, 0\right), N\left(x_{2}, 0\right)$ $\left(0<x_{1}<x_{2}\right)$, and passes through the point $A(0,1)$. A line $l$ passing through point $A$ intersects the $x$-axis at point $C$ and the parabola at point $B$ (different from point $A$), satisfying that $\triangle C A N$ is an isosceles right triangle, and $S_{\triangle B M N}=\frac{5}{2} S_{\triangle A M N}$. Find the equation of the parabola.
2. From the conditions, we know that the parabola opens upwards, and its two intersections with the $x$-axis are to the right of the $y$-axis.

Since $\triangle C A N$ is an isosceles right triangle, point $C$ is on the left side of the $x$-axis, and
$\angle C A N=90^{\circ}, \angle A C N=45^{\circ}$.
Thus, $C(-1,0), N(1,0)$.
Therefore, the equation of line $l$ is $y=x+1$.
Let $B\left(x_{3}, y_{3}\right)$.
Given $S_{\triangle B M N}=\frac{5}{2} S_{\triangle M M N}$, we know $y_{3}=\frac{5}{2}$.
Thus, $x_{3}=\frac{3}{2}$, i.e., $B\left(\frac{3}{2}, \frac{5}{2}\right)$.
In summary, the parabola passes through points $A(0,1) 、 B\left(\frac{3}{2}, \frac{5}{2}\right)$ 、 $N(1,0)$.
Therefore, $\left\{\begin{array}{l}1=c, \\ \frac{5}{2}=\frac{9}{4} a+\frac{3}{2} b+c, \\ 0=a+b+c\end{array} \Rightarrow\left\{\begin{array}{l}a=4, \\ b=-5, \\ c=1 .\end{array}\right.\right.$
Hence, the equation of the parabola is $y=4 x^{2}-5 x+1$.