6x^2+17<0 inequation

Given the inequality:
$$6 x^{2} + 17 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$6 x^{2} + 17 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 6$$
$$b = 0$$
$$c = 17$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 6 \cdot 4 \cdot 17 + 0^{2} = -408$$
Because D<0, then the equation
has no real roots,
but complex roots is exists.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = \frac{\sqrt{102} i}{6}$$
Simplify
$$x_{2} = - \frac{\sqrt{102} i}{6}$$
Simplify
$$x_{1} = \frac{\sqrt{102} i}{6}$$
$$x_{2} = - \frac{\sqrt{102} i}{6}$$
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
$$x_0 = 0$$
$$6 \cdot 0^{2} + 17 < 0$$

17 < 0

but

17 > 0

so the inequality has no solutions