Given the inequality:
$$\left(- 2 x^{2} + 11 x\right) + 12 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 2 x^{2} + 11 x\right) + 12 = 0$$
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
$$a = -2$$
$$b = 11$$
$$c = 12$$
, then
D = b^2 - 4 * a * c =
(11)^2 - 4 * (-2) * (12) = 217
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{11}{4} - \frac{\sqrt{217}}{4}$$
$$x_{2} = \frac{11}{4} + \frac{\sqrt{217}}{4}$$
$$x_{1} = \frac{11}{4} - \frac{\sqrt{217}}{4}$$
$$x_{2} = \frac{11}{4} + \frac{\sqrt{217}}{4}$$
$$x_{1} = \frac{11}{4} - \frac{\sqrt{217}}{4}$$
$$x_{2} = \frac{11}{4} + \frac{\sqrt{217}}{4}$$
This roots
$$x_{1} = \frac{11}{4} - \frac{\sqrt{217}}{4}$$
$$x_{2} = \frac{11}{4} + \frac{\sqrt{217}}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{11}{4} - \frac{\sqrt{217}}{4}\right) + - \frac{1}{10}$$
=
$$\frac{53}{20} - \frac{\sqrt{217}}{4}$$
substitute to the expression
$$\left(- 2 x^{2} + 11 x\right) + 12 \geq 0$$
$$\left(11 \left(\frac{53}{20} - \frac{\sqrt{217}}{4}\right) - 2 \left(\frac{53}{20} - \frac{\sqrt{217}}{4}\right)^{2}\right) + 12 \geq 0$$
2
/ _____\ _____
823 |53 \/ 217 | 11*\/ 217 >= 0
--- - 2*|-- - -------| - ----------
20 \20 4 / 4
but
2
/ _____\ _____
823 |53 \/ 217 | 11*\/ 217 < 0
--- - 2*|-- - -------| - ----------
20 \20 4 / 4
Then
$$x \leq \frac{11}{4} - \frac{\sqrt{217}}{4}$$
no execute
one of the solutions of our inequality is:
$$x \geq \frac{11}{4} - \frac{\sqrt{217}}{4} \wedge x \leq \frac{11}{4} + \frac{\sqrt{217}}{4}$$
_____
/ \
-------•-------•-------
x1 x2