find the equation of the normal to the curve f(x)=7x-2x^2 at the point where x=6

First we find the equation for tangent line;

\(f'(x)=-4x+7\\ \text{Gradient of tangent line}=f'(6)=7-24 = -17\)

When x = 6, f(x)=42 - 2(36)= -30

\(y-(-30)=-17(x-6)\\ y=-17x+72\)

There we go!!

Then we find the gradient for normal.

Negative reciprocal of -17 is 1/17:
Gradient of normal is 1/17

Equation for the normal:

 \(y-(-30)=\dfrac{x-6}{17}\\ y=\dfrac{x-516}{17}\)

Did I make a mistake? This is the first time I know what is normal line. :D