Find the tangent to the curve y=7x3-2x2 at the point x=2

The correct option is "76" (which represents the slope of the tangent line to the curve at the given point).

To find the slope of the tangent to the curve at a specific point, we need to find the derivative of the curve's equation with respect to $x$ and then evaluate it at the given value of $x$.

The equation of the curve is given by:
$$y=7x^3-2x^2$$

First, we find the first derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$, which represents the slope of the tangent line at any point $x$. Using the power rule of differentiation ($\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$):
$$\frac{dy}{dx}=\frac{d}{dx}\left(7x^3-2x^2\right)$$
$$\frac{dy}{dx}=7\cdot \left(3x^2\right)-2\cdot \left(2x\right)$$
$$\frac{dy}{dx}=21x^2-4x$$

Next, we substitute $x=2$ into the derivative to calculate the slope of the tangent at this particular point:
$${\left.\frac{dy}{dx}\right|}_{x=2}=21{\left(2\right)}^2-4\left(2\right)$$
$${\left.\frac{dy}{dx}\right|}_{x=2}=21\left(4\right)-8$$
$${\left.\frac{dy}{dx}\right|}_{x=2}=84-8$$
$${\left.\frac{dy}{dx}\right|}_{x=2}=76$$

Therefore, the slope of the tangent to the curve at $x=2$ is $76$.