We have to solve for values of x that satisfy (x^2 -3x -4) < 0

(x^2 -3x -4) < 0

=> x^2 - 4x + x - 4 < 0

=> x(x - 4) +1(x - 4) < 0

=> (x + 1)(x - 4) < 0

If the above condition has to be true either of the terms should be negative and the other positive

=> x + 1 < 0 and x - 4 >=0

=> x < -1 and x >=4

This gives no valid values.

x + 1 >=0 and x - 4 < 0

=> x >= -1 and x < 4

**This gives the values of x in the set [-1 , 4)**

Given the inequality:

(x^2 - 3x -4) < 0

First we will factor .

==> (x-4)(x+1) < 0

Since the product is negative, then the terms should have different signs.

==> Then we have two cases:

(x-4) < 0 and (x+1) > 0

==> x < 4 and x > -1

==> -1 < x < 4

==> x = (-1, 4)

OR:

(x-4) > 0 and x+1 < 0

==> x > 4 and x < -1 ( no solution).

**Then the solution is that x belongs to the interval ( -1, 4).**