Question

How do you solve `6c^2 -72 =11c`?

Answer 1

`- 8/3 and 9/2`

Explanation:

`y = 6c^2 - 11c - 72 = 0`
Use the new Transforming Method (Socratic Search).
Transformed equation: `y' = c^2 - 11c - 432 = 0`
The 2 roots of y' have opposite signs because ac < 0.
Compose factor pairs of (ac = - 432) with a calculator -->
... (-12, 36)(-16, 27). This last sum is (11 = -b).
Therefor, the 2 real roots of y' are: -16 and 27.
Back to original equation, the 2 real roots are:
`c1 = -16/a = -16/6 = -8/3` and `c2 = 27/a = 27/6 = 9/2`

Reminder:
Sum of 2 real roots --> -b
Product of 2 real roots --> c