(2x+5x-6)^2≥64 inequation

Given the inequality:
$$\left(\left(2 x + 5 x\right) - 6\right)^{2} \geq 64$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\left(2 x + 5 x\right) - 6\right)^{2} = 64$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\left(\left(2 x + 5 x\right) - 6\right)^{2} = 64$$
to
$$\left(\left(2 x + 5 x\right) - 6\right)^{2} - 64 = 0$$
Expand the expression in the equation
$$\left(\left(2 x + 5 x\right) - 6\right)^{2} - 64 = 0$$
We get the quadratic equation
$$49 x^{2} - 84 x - 28 = 0$$
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
$$a = 49$$
$$b = -84$$
$$c = -28$$
, then

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

(-84)^2 - 4 * (49) * (-28) = 12544

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

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

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

or
$$x_{1} = 2$$
$$x_{2} = - \frac{2}{7}$$
$$x_{1} = 2$$
$$x_{2} = - \frac{2}{7}$$
$$x_{1} = 2$$
$$x_{2} = - \frac{2}{7}$$
This roots
$$x_{2} = - \frac{2}{7}$$
$$x_{1} = 2$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{2}{7} + - \frac{1}{10}$$
=
$$- \frac{27}{70}$$
substitute to the expression
$$\left(\left(2 x + 5 x\right) - 6\right)^{2} \geq 64$$
$$\left(-6 + \left(\frac{\left(-27\right) 5}{70} + \frac{\left(-27\right) 2}{70}\right)\right)^{2} \geq 64$$

7569      
---- >= 64
100       

one of the solutions of our inequality is:
$$x \leq - \frac{2}{7}$$

 _____           _____          
      \         /
-------•-------•-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq - \frac{2}{7}$$
$$x \geq 2$$