You need to solve the following quadratic inequality, such that:

`2(x - 1/2)(x + 2) < 0 => (x - 1/2)(x + 2) < 0 `

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

You need to attach quadratic equation, such that:

`2x^2 + 3x - 2 = 0`

Using quadratic formula yields:

`x_(1,2) = (-3 +- sqrt(9 + 16))/4 => x_(1,2) = (-3 +- sqrt25)/4`

`x_(1,2) = (-3 +- 5)/4 => x_1 = 1/2 ; x_2 = -2`

**Hence, evaluating where the quadratic is negative yields that it stays negative between -`2` and `1/2` .**

We'll divide both sides by 2:

Since the value is positive, the inequality keeps it's sense:

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

We'll conclude that a product is negative if the factors are of opposite sign.

We'll discuss 2 cases:

1) (x - 1/2) < 0

and

(x+2) > 0

We'll solve the first inequality. For this reason, we'll isolate x to the left side.

x < 1/2

We'll solve the 2nd inequality:

(x+2) > 0

We'll subtract 2 both sides:

x > -2

The common solution of the first system of inequalities is the interval (-2 , 1/2).

We'll solve the second system of inequalities:

2) (x-1/2) > 0

and

(x+2) < 0

x-1/2 > 0

x > 1/2

(x+2) < 0

x < -2

Therefore, the complete solution is the solution from the first system of inequalities, the opened interval (-2 , 1/2).