Given the inequality:
$$\left(x + 6\right) \left(5 x - 2\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x + 6\right) \left(5 x - 2\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(x + 6\right) \left(5 x - 2\right) = 0$$
We get the quadratic equation
$$5 x^{2} + 28 x - 12 = 0$$
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
$$a = 5$$
$$b = 28$$
$$c = -12$$
, then
D = b^2 - 4 * a * c =
(28)^2 - 4 * (5) * (-12) = 1024
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{2}{5}$$
$$x_{2} = -6$$
$$x_{1} = \frac{2}{5}$$
$$x_{2} = -6$$
$$x_{1} = \frac{2}{5}$$
$$x_{2} = -6$$
This roots
$$x_{2} = -6$$
$$x_{1} = \frac{2}{5}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-6 + - \frac{1}{10}$$
=
$$- \frac{61}{10}$$
substitute to the expression
$$\left(x + 6\right) \left(5 x - 2\right) > 0$$
$$\left(- \frac{61}{10} + 6\right) \left(\frac{\left(-61\right) 5}{10} - 2\right) > 0$$
13/4 > 0
one of the solutions of our inequality is:
$$x < -6$$
_____ _____
\ /
-------ο-------ο-------
x2 x1Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -6$$
$$x > \frac{2}{5}$$