Given x1 = -2.5 and x2 = 4.

The quadratic curve passes through $(-2.5, 0)$ and $(4, 0)$. These are the x-intercepts (roots) of the equation.
A quadratic equation with roots $x_1$ and $x_2$ can be written in the form $y = a(x - x_1)(x - x_2)$.
Given $x_1 = -2.5$ and $x_2 = 4$.

Substitute the roots into the equation:

$$y = a(x - (-2.5))(x - 4)$$ $$y = a(x + 2.5)(x - 4)$$

First, expand the product of the binomials:

$$(x + 2.5)(x - 4) = x(x - 4) + 2.5(x - 4)$$ $$= x^2 - 4x + 2.5x - 10$$ $$= x^2 - 1.5x - 10$$

Now substitute this back into the equation:

$$y = a(x^2 - 1.5x - 10)$$

The equation is in the form $y = ax^2 + bx + c$.
From our expanded form, $y = ax^2 - 1.5ax - 10a$.
For $b = -1.5a$ and $c = -10a$ to be integers, $a$ must be chosen carefully.
The coefficient $-1.5$ can be written as $-\frac{3}{2}$.
So, $y = a\left(x^2 - \frac{3}{2}x - 10\right)$.
To make all coefficients integers, $a$ must be a multiple of the denominator of the fraction, which is 2. The smallest positive integer value for $a$ is 2.

Let $a = 2$:

$$y = 2\left(x^2 - \frac{3}{2}x - 10\right)$$ $$y = 2x^2 - 2\left(\frac{3}{2}x\right) - 2(10)$$ $$y = 2x^2 - 3x - 20$$

In this equation, $a=2$, $b=-3$, and $c=-20$, which are all integers.