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(z-7)*(4z+2)>0 inequation

A inequation with variable

The solution

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(z - 7)*(4*z + 2) > 0
$$\left(z - 7\right) \left(4 z + 2\right) > 0$$
(z - 7)*(4*z + 2) > 0
Detail solution
Given the inequality:
$$\left(z - 7\right) \left(4 z + 2\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(z - 7\right) \left(4 z + 2\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(z - 7\right) \left(4 z + 2\right) = 0$$
We get the quadratic equation
$$4 z^{2} - 26 z - 14 = 0$$
This equation is of the form
a*z^2 + b*z + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$z_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 4$$
$$b = -26$$
$$c = -14$$
, then
D = b^2 - 4 * a * c = 

(-26)^2 - 4 * (4) * (-14) = 900

Because D > 0, then the equation has two roots.
z1 = (-b + sqrt(D)) / (2*a)

z2 = (-b - sqrt(D)) / (2*a)

or
$$z_{1} = 7$$
$$z_{2} = - \frac{1}{2}$$
$$z_{1} = 7$$
$$z_{2} = - \frac{1}{2}$$
$$z_{1} = 7$$
$$z_{2} = - \frac{1}{2}$$
This roots
$$z_{2} = - \frac{1}{2}$$
$$z_{1} = 7$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$z_{0} < z_{2}$$
For example, let's take the point
$$z_{0} = z_{2} - \frac{1}{10}$$
=
$$- \frac{1}{2} + - \frac{1}{10}$$
=
$$- \frac{3}{5}$$
substitute to the expression
$$\left(z - 7\right) \left(4 z + 2\right) > 0$$
$$\left(-7 + - \frac{3}{5}\right) \left(\frac{\left(-3\right) 4}{5} + 2\right) > 0$$
76    
-- > 0
25    

one of the solutions of our inequality is:
$$z < - \frac{1}{2}$$
 _____           _____          
      \         /
-------ο-------ο-------
       z2      z1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$z < - \frac{1}{2}$$
$$z > 7$$
Solving inequality on a graph
Rapid solution [src]
Or(And(-oo < z, z < -1/2), And(7 < z, z < oo))
$$\left(-\infty < z \wedge z < - \frac{1}{2}\right) \vee \left(7 < z \wedge z < \infty\right)$$
((-oo < z)∧(z < -1/2))∨((7 < z)∧(z < oo))
Rapid solution 2 [src]
(-oo, -1/2) U (7, oo)
$$z\ in\ \left(-\infty, - \frac{1}{2}\right) \cup \left(7, \infty\right)$$
z in Union(Interval.open(-oo, -1/2), Interval.open(7, oo))