Given the inequality:
$$\left(6 x^{2} - 7 x\right) - 20 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(6 x^{2} - 7 x\right) - 20 = 0$$
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
$$a = 6$$
$$b = -7$$
$$c = -20$$
, then
D = b^2 - 4 * a * c =
(-7)^2 - 4 * (6) * (-20) = 529
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{5}{2}$$
$$x_{2} = - \frac{4}{3}$$
$$x_{1} = \frac{5}{2}$$
$$x_{2} = - \frac{4}{3}$$
$$x_{1} = \frac{5}{2}$$
$$x_{2} = - \frac{4}{3}$$
This roots
$$x_{2} = - \frac{4}{3}$$
$$x_{1} = \frac{5}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{4}{3} + - \frac{1}{10}$$
=
$$- \frac{43}{30}$$
substitute to the expression
$$\left(6 x^{2} - 7 x\right) - 20 < 0$$
$$-20 + \left(- \frac{\left(-43\right) 7}{30} + 6 \left(- \frac{43}{30}\right)^{2}\right) < 0$$
59
-- < 0
25
but
59
-- > 0
25
Then
$$x < - \frac{4}{3}$$
no execute
one of the solutions of our inequality is:
$$x > - \frac{4}{3} \wedge x < \frac{5}{2}$$
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