x^2 + x - 6 =< 0

First let us factor:

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

Then we have two cases:

case (1):

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

=> x =< -3 and x >=2

The solution = empty set.

Case (2):

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

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

==> -3 =< x =< 2

**==> x = { -3, -2, -1, 0, 1, 2}**

For the integer solutions of the inequality x^2 + x - 6 = < 0.

We have x^2 + x - 6 = < 0

=> x^2 + 3x - 2x -6 = < 0

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

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

For an equality we have the values x = 2 Aden x = -3 .

For values of the expression less than 0,

either x -2 < 0 and x+3 >0

=> x < 2 and x > -3

or x -2 >0 and x+3 <0

=> x > 2 and x < -3 which is not possible.

Therefore x can be x >= -3 and x<= 2

**The integer solutions are -3, -2, -1, 0, 1, and 2**

x^2+x-6 < 0

To find integer solution:

Factor the LHS of the inequality:

x^2+3x -2x-6 < 0

x(x+3) -2(x+3) < 0

(x+3)(x-2) < 0

Therefore x >-3 and x <2 : Th solution inintegers: {-2,-1 ,0, 1}

We'll determine first the roots of the quadratic equation:

x^2+x-6 = 0

We'll apply the quadratic formula:

x1 = [-1 + sqrt(1+24)]/2

x1 = (-1+5)/2

x1 = 2

x2 = (-1-5)/2

x2 = -3

The expression x^2+x-6 is negative or zero between the roots, for x having values over teh interval [-3 ; 2].

Now, we'll extract from the interval [-3 ; 2] only the integer values.

**The integer solutions of the inequality are:**

**{-3 , -2 , -1 , 0 , 1 , 2}.**