Find the equation of all the lines having slope 0 which are tangent to the curve y=6x2-7x

The correct option is -(49/24).

We are asked to find the equation of all lines with a slope of $0$ that are tangent to the curve given by:
$y=6x^2-7x$

First, we recall that the slope of the tangent line to a curve $y=f\left(x\right)$ at any point is given by the derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$. Let us calculate this derivative using the power rule of differentiation:
$\frac{dy}{dx}=\frac{d}{dx}\left(6x^2-7x\right)$
$\frac{dy}{dx}=12x-7$

Since the problem states that the tangent line has a slope of $0$, we set the derivative equal to $0$ to find the x-coordinate of the point of tangency:
$12x-7=0$
$12x=7$
$x=\frac{7}{12}$

Now, we find the corresponding y-coordinate by substituting the value of $x$ back into the original curve's equation:
$y=6{\left(\frac{7}{12}\right)}^2-7\left(\frac{7}{12}\right)$
$y=6\left(\frac{49}{144}\right)-\frac{49}{12}$
Since $144÷6=24$, the first term simplifies as follows:
$y=\frac{49}{24}-\frac{49}{12}$

To subtract these fractions, we find a common denominator, which is $24$:
$y=\frac{49}{24}-\frac{49\times 2}{24}$
$y=\frac{49}{24}-\frac{98}{24}$
$y=-\frac{49}{24}$

The equation of a horizontal line (having slope $m=0$) passing through a point $(x_1,y_1)$ is simply given by $y=y_1$. Therefore, the equation of the tangent line is:
$y=-\frac{49}{24}$