Find an equation for the line tangent to $y=2-6 x^{2}$ at $(-5,-148)$.
The equation for the line tangent to $y=2-6 x^{2}$ at $(-5,-148)$ is $y=$

Step 1 ：Given the function \(y=2-6 x^{2}\), we want to find the equation of the line tangent to this function at the point \((-5,-148)\).

Step 2 ：The equation of a tangent line to a function at a given point can be found using the formula \(y = mx + b\), where \(m\) is the slope of the tangent line and \(b\) is the y-intercept.

Step 3 ：The slope of the tangent line is the derivative of the function evaluated at the given point. So, first we need to find the derivative of the function \(y=2-6 x^{2}\).

Step 4 ：The derivative of the function \(y=2-6 x^{2}\) is \(-12x\).

Step 5 ：Evaluating this derivative at \(x=-5\) gives us a slope \(m=60\).

Step 6 ：We can find the y-intercept by substituting the given point into the equation of the line. This gives us the equation \(-148 = 60*(-5) + b\).

Step 7 ：Solving this equation for \(b\) gives us \(b=152\).

Step 8 ：Thus, the equation of the line tangent to \(y=2-6 x^{2}\) at \((-5,-148)\) is \(y=60x+152\).

Step 9 ：\(\boxed{y=60x+152}\) is the final answer.