6-7x-3x^2>0 inequation

Given the inequality:
$$- 3 x^{2} + \left(6 - 7 x\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- 3 x^{2} + \left(6 - 7 x\right) = 0$$
Solve:
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
$$a = -3$$
$$b = -7$$
$$c = 6$$
, then

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

(-7)^2 - 4 * (-3) * (6) = 121

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

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

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

or
$$x_{1} = -3$$
$$x_{2} = \frac{2}{3}$$
$$x_{1} = -3$$
$$x_{2} = \frac{2}{3}$$
$$x_{1} = -3$$
$$x_{2} = \frac{2}{3}$$
This roots
$$x_{1} = -3$$
$$x_{2} = \frac{2}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-3 + - \frac{1}{10}$$
=
$$- \frac{31}{10}$$
substitute to the expression
$$- 3 x^{2} + \left(6 - 7 x\right) > 0$$
$$- 3 \left(- \frac{31}{10}\right)^{2} + \left(6 - \frac{\left(-31\right) 7}{10}\right) > 0$$

-113     
----- > 0
 100     

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

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       x1      x2