4x^2-20x+25<=0 inequation

Given the inequality:
$$\left(4 x^{2} - 20 x\right) + 25 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(4 x^{2} - 20 x\right) + 25 = 0$$
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
$$a = 4$$
$$b = -20$$
$$c = 25$$
, then

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

(-20)^2 - 4 * (4) * (25) = 0

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

x = -b/2a = --20/2/(4)

$$x_{1} = \frac{5}{2}$$
$$x_{1} = \frac{5}{2}$$
$$x_{1} = \frac{5}{2}$$
This roots
$$x_{1} = \frac{5}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{5}{2}$$
=
$$\frac{12}{5}$$
substitute to the expression
$$\left(4 x^{2} - 20 x\right) + 25 \leq 0$$
$$\left(- \frac{12 \cdot 20}{5} + 4 \left(\frac{12}{5}\right)^{2}\right) + 25 \leq 0$$

1/25 <= 0

but

1/25 >= 0

Then
$$x \leq \frac{5}{2}$$
no execute
the solution of our inequality is:
$$x \geq \frac{5}{2}$$

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