(2x+3)(2-x)>3 inequation

Given the inequality:
$$\left(2 - x\right) \left(2 x + 3\right) > 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(2 - x\right) \left(2 x + 3\right) = 3$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\left(2 - x\right) \left(2 x + 3\right) = 3$$
to
$$\left(2 - x\right) \left(2 x + 3\right) - 3 = 0$$
Expand the expression in the equation
$$\left(2 - x\right) \left(2 x + 3\right) - 3 = 0$$
We get the quadratic equation
$$- 2 x^{2} + x + 3 = 0$$
This equation is of the form

a*x^2 + b*x + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -2$$
$$b = 1$$
$$c = 3$$
, then

D = b^2 - 4 * a * c = 

(1)^2 - 4 * (-2) * (3) = 25

Because D > 0, then the equation has two roots.

x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = -1$$
$$x_{2} = \frac{3}{2}$$
$$x_{1} = -1$$
$$x_{2} = \frac{3}{2}$$
$$x_{1} = -1$$
$$x_{2} = \frac{3}{2}$$
This roots
$$x_{1} = -1$$
$$x_{2} = \frac{3}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$\left(2 - x\right) \left(2 x + 3\right) > 3$$
$$\left(2 - - \frac{11}{10}\right) \left(\frac{\left(-11\right) 2}{10} + 3\right) > 3$$

62    
-- > 3
25    

Then
$$x < -1$$
no execute
one of the solutions of our inequality is:
$$x > -1 \wedge x < \frac{3}{2}$$

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       x1      x2