3x^2+7x+2<=0 inequation

Given the inequality:
$$3 x^{2} + 7 x + 2 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$3 x^{2} + 7 x + 2 = 0$$
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
$$a = 3$$
$$b = 7$$
$$c = 2$$
, then

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

(7)^2 - 4 * (3) * (2) = 25

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

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

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

or
$$x_{1} = - \frac{1}{3}$$
Simplify
$$x_{2} = -2$$
Simplify
$$x_{1} = - \frac{1}{3}$$
$$x_{2} = -2$$
$$x_{1} = - \frac{1}{3}$$
$$x_{2} = -2$$
This roots
$$x_{2} = -2$$
$$x_{1} = - \frac{1}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-2 - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$3 x^{2} + 7 x + 2 \leq 0$$
$$7 \left(- \frac{21}{10}\right) + 2 + 3 \left(- \frac{21}{10}\right)^{2} \leq 0$$

 53     
--- <= 0
100     

but

 53     
--- >= 0
100     

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

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        /     \  
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       x_2      x_1