We need to find x that satisfies x^2 - x - 6 > 0

x^2 - x - 6 > 0

=> x^2 – 3x + 2x – 6 > 0

=> x(x – 3) + 2(x – 3) > 0

=> (x + 2) (x – 3) > 0

...

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We need to find x that satisfies x^2 - x - 6 > 0

x^2 - x - 6 > 0

=> x^2 – 3x + 2x – 6 > 0

=> x(x – 3) + 2(x – 3) > 0

=> (x + 2) (x – 3) > 0

For this to be valid either both (x + 2) and (x – 3) should be greater than 0 or both (x + 2) and (x – 3) should be less than 0.

In the first case (x + 2) > 0 and (x – 3) > 0

=> x > -2 and x > 3

x > 3 satisfies both the conditions.

In the second case (x + 2) < 0 and (x – 3) < 0

=> x < -2 and x < 3

x < -2 satisfies both the conditions.

This gives us the values of x as either greater than 3 or less than -2.

**The required values of x lie in (-inf, - 2) U (3, inf)**

x^2- x - 6 > 0

First we will factor the left side.

==> (x-3)(x+2) > 0

Now we have a product of two terms.

In order for the product to be positive, then both terms should be positive or both terms should be negative.

Then we will write:

x-3 > 0 AND x+2 > 0

==> x > 3 AND x > -2

==> x = ( 3, inf ) n (-2, inf) = (3,inf) ..........(1)

Now for the other option, both terms are negative.

==> x-3 < 0 AND x+ 2 < 0

==> x < 3 AND x < -2

==> x = ( -inf, 3) n ( -inf , -2) = ( -inf, -2)............(2)

Then, from (1) and (2) we have two solutions:

**==> x = ( -inf , -2) U ( 3, inf)**