-4x^2+20x>25 inequation

Given the inequality:
$$- 4 x^{2} + 20 x > 25$$
To solve this inequality, we must first solve the corresponding equation:
$$- 4 x^{2} + 20 x = 25$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$- 4 x^{2} + 20 x = 25$$
to
$$\left(- 4 x^{2} + 20 x\right) - 25 = 0$$
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
$$a = -4$$
$$b = 20$$
$$c = -25$$
, 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:
$$x_{0} < x_{1}$$
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
$$- 4 x^{2} + 20 x > 25$$
$$- 4 \left(\frac{12}{5}\right)^{2} + \frac{12 \cdot 20}{5} > 25$$

624     
--- > 25
 25     

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

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