We have to find the solutions of 3x^2 + 6x < 9

3x^2 + 6x < 9

=> 3x^2 + 6x - 9 < 0

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

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

=> x(x + 3) -1 (x + 3) < 0

=> (x - 1)(x + 3) < 0

For (x - 1)(x + 3) to be less than 0 , either of them should be less than 0

=> x - 1 < 0 and x + 3 > 0

=> x < 1 and x > -3

This gives a set of values as (-3 , 1)

x - 1 > 0 and x + 3 < 0

=> x > 1 and x < -3 gives no solutions.

As x is an integer (-3 , 1) implies (-2 , -1 , 0)

**So x can have the values (-2 , -1 , 0).**

Given the inequality: 3x^2 + 6x < 9

We need to find x such that x E Z

First we will solve the inequality.

==> 3x^2 + 6x - 9 < 0

Now we will factor 3.

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

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

Now we will find the roots.

==> (x+3)(x-1) < 0

==> x+ 3 < 0 and x-1 > 0

==> x < -3 and x > 1 ( impossible)

==> (x+3) > 0 and (x-1) <0

==> x > -3 and x < 1

==> -3 < x < 1

Then the values of x are between the numbers -3 and 1

But x is an integer.

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