How can I find the values of x that satisfy x^2 - 4x -12 > 0

To find the solutions for the equation x^2-4x+12 >
0.

We first factorise the left side  expression
x^2-4x-12.

x^2-4x+12 =
x^2-6x+2x-12

x^2-4x+12 = x(x-6)
+2(x-6)

x^2-4x+12 = (x-6)
(x+2).

Therefore  x^2-4x-12> 0 implies (x+2)(x-6) >
0.

Therefore both factors should be of the same
sign.

Therefore x+2 < 0 and x-6 < 0
,

Or (x+2)> 0 and (x-6) >
0.

This is possinble only when  x < -2   Or  x
> 6 , the x^2-4x-12 > 0.

Or x must be
in the interval (-infinity -2) Or in the interval (6  infinity).