2x²-3x+10<0 inequation

Given the inequality:
$$\left(2 x^{2} - 3 x\right) + 10 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(2 x^{2} - 3 x\right) + 10 = 0$$
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
$$a = 2$$
$$b = -3$$
$$c = 10$$
, then

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

(-3)^2 - 4 * (2) * (10) = -71

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
$$x_{1} = \frac{3}{4} + \frac{\sqrt{71} i}{4}$$
$$x_{2} = \frac{3}{4} - \frac{\sqrt{71} i}{4}$$
$$x_{1} = \frac{3}{4} + \frac{\sqrt{71} i}{4}$$
$$x_{2} = \frac{3}{4} - \frac{\sqrt{71} i}{4}$$
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

$$\left(2 \cdot 0^{2} - 0 \cdot 3\right) + 10 < 0$$

10 < 0

but

10 > 0

so the inequality has no solutions