Find the coordinates of the points on the curve $y = 2x^3 - 9x^2 - 12x + 7$ where the gradient is 12

I've tried multiple times so I must be doing something wrong. I differentiated to get $dy/dx = 6x^2 - 18x - 12$ and then I set that equal to $12$ and rearranged/factorised to get $x = 4$ and $x = -1$.

The answer is (according to the back of the book): $(-1,8)$ and $(4,-57)$.

You are on the right track. Once you have found the x-values ($-1$ and $4$), plug those (separately) into the original equation and solve for the corresponding y-coordinates.
you must solve the equation $$f'(x)=6x^2-18x-12=12$$ for $x$