x^2+2x-15<=0 inequation

Given the inequality:
$$x^{2} + 2 x - 15 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$x^{2} + 2 x - 15 = 0$$
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
$$a = 1$$
$$b = 2$$
$$c = -15$$
, then

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

(2)^2 - 4 * (1) * (-15) = 64

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

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

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

or
$$x_{1} = 3$$
Simplify
$$x_{2} = -5$$
Simplify
$$x_{1} = 3$$
$$x_{2} = -5$$
$$x_{1} = 3$$
$$x_{2} = -5$$
This roots
$$x_{2} = -5$$
$$x_{1} = 3$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-5 - \frac{1}{10}$$
=
$$- \frac{51}{10}$$
substitute to the expression
$$x^{2} + 2 x - 15 \leq 0$$
$$\left(-1\right) 15 + 2 \left(- \frac{51}{10}\right) + \left(- \frac{51}{10}\right)^{2} \leq 0$$

 81     
--- <= 0
100     

but

 81     
--- >= 0
100     

Then
$$x \leq -5$$
no execute
one of the solutions of our inequality is:
$$x \geq -5 \wedge x \leq 3$$

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        /     \  
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       x_2      x_1