2x^2 + x -3 < 0

First let us factor:

(2x +3)(x-1) < 0

Now we have a product of two terms.

The product is a negative number.

Then one of the terms should be negative:

==> (2x+3) < 0 AND (x-1) > 0

==> x < -3/2 AND x > 1 (impossible)

OR

( 2x+3) > 0 AND (x-1) < 0

==> x > -3/2 AND x < 1

==> -3/2 < x < 1

**==> x = ( -3/2, 1)**

We are given the following inequation to solve 2x^2 + x - 3 < 0

2x^2 + x - 3 < 0

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

=> x ( 2x +3) -1 (2x + 3) <0

=> (x - 1)(2x + 3) <0

Now as ( x - 1)(2x + 3) is less than 0, either (x - 1) < 0 and (2x + 3) >0 or (x - 1) > 0 and (2x + 3) < 0.

For the first case:

(x - 1) < 0 and (2x + 3) >0

=> x < 1 and x > -3/2

Therefore x takes values less than 1 and greater than -3/2 , or x lies in (-3/2 , 1)

For the second case:

(x - 1) > 0 and (2x + 3) < 0

=> x > 1 and x < -3/2

This implies x has to be simultaneously greater than 1 and less than -3/2, which is not possible.

So we cannot obtain any values for x here.

**Therefore the possible values for x are ( -3/2 , 1)**