(x+5)^2 inequation

Given the inequality:
$$\left(x + 5\right)^{2} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x + 5\right)^{2} = 0$$
Solve:
Expand the expression in the equation
$$\left(x + 5\right)^{2} = 0$$
We get the quadratic equation
$$x^{2} + 10 x + 25 = 0$$
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
$$a = 1$$
$$b = 10$$
$$c = 25$$
, then

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

(10)^2 - 4 * (1) * (25) = 0

Because D = 0, then the equation has one root.

x = -b/2a = -10/2/(1)

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

1/100 > 0

the solution of our inequality is:
$$x < -5$$

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