Find the derivative of y = (x2−3x+5)(2x−7) using the product rule.

Let $u = x^2-3x+5$.
Then, the derivative of $u$ with respect to $x$ is:

$$\frac{du}{dx} = \frac{d}{dx}(x^2-3x+5) = 2x-3$$

Let $v = 2x-7$.
Then, the derivative of $v$ with respect to $x$ is:

$$\frac{dv}{dx} = \frac{d}{dx}(2x-7) = 2$$

$$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$

Substitute the expressions for $u, v, \frac{du}{dx},$ and $\frac{dv}{dx}$:

$$\frac{dy}{dx} = (x^2-3x+5)(2) + (2x-7)(2x-3)$$

First, expand the first term:

$$2(x^2-3x+5) = 2x^2 - 6x + 10$$

Next, expand the second term:

$$(2x-7)(2x-3) = 2x(2x-3) - 7(2x-3)$$ $$= 4x^2 - 6x - 14x + 21$$ $$= 4x^2 - 20x + 21$$

$$\frac{dy}{dx} = (2x^2 - 6x + 10) + (4x^2 - 20x + 21)$$

Combine like terms:

$$\frac{dy}{dx} = (2x^2 + 4x^2) + (-6x - 20x) + (10 + 21)$$ $$\frac{dy}{dx} = 6x^2 - 26x + 31$$

The derivative of $(x^2-3x+5)(2x-7)$ is $\boxed{6x^2 - 26x + 31}$.