Let f(x) = ax^3 + 6x^2 + bx +4. determine the constants a and b so that f has a relative minimum at x = -1 and a relative maximum at x = 2. show workings

F'(x) = 3ax^2 + 12x + b (1)
Applying the formula for solving any quadratic leads to two equations:
(-12 + sq rt(144 - 12ab))/6a = 2 (2)
(-12 - sq rt(144 - 12ab))/6a = -1 (3)
Adding equations (2) and (3) , we get:
-24/6a = 1
a = -4
Plugging the value a = -4 into equation (2) and solving for b, we get:
b = 24.

The answer is: a = -4, b = 24.

So confused right here i need it to be more clearer