18x^2-6x+4<0 inequation

Given the inequality:
$$\left(18 x^{2} - 6 x\right) + 4 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(18 x^{2} - 6 x\right) + 4 = 0$$
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
$$a = 18$$
$$b = -6$$
$$c = 4$$
, then

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

(-6)^2 - 4 * (18) * (4) = -252

Because D<0, then the equation
has no real roots,
but complex roots is exists.

x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = \frac{1}{6} + \frac{\sqrt{7} i}{6}$$
$$x_{2} = \frac{1}{6} - \frac{\sqrt{7} i}{6}$$
$$x_{1} = \frac{1}{6} + \frac{\sqrt{7} i}{6}$$
$$x_{2} = \frac{1}{6} - \frac{\sqrt{7} i}{6}$$
Exclude the complex solutions:
This equation has no roots,
this inequality is executed for any x value or has no solutions
check it
subtitute random point x, for example

x0 = 0

$$\left(18 \cdot 0^{2} - 0 \cdot 6\right) + 4 < 0$$

4 < 0

but

4 > 0

so the inequality has no solutions