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2x^2-3x+5>=0

2x^2-3x+5>=0 inequation

A inequation with variable

The solution

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   2               
2*x  - 3*x + 5 >= 0
$$2 x^{2} - 3 x + 5 \geq 0$$
2*x^2 - 3*x + 5 >= 0
Detail solution
Given the inequality:
$$2 x^{2} - 3 x + 5 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 x^{2} - 3 x + 5 = 0$$
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
$$a = 2$$
$$b = -3$$
$$c = 5$$
, then
D = b^2 - 4 * a * c = 

(-3)^2 - 4 * (2) * (5) = -31

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
$$x_{1} = \frac{3}{4} + \frac{\sqrt{31} i}{4}$$
Simplify
$$x_{2} = \frac{3}{4} - \frac{\sqrt{31} i}{4}$$
Simplify
$$x_{1} = \frac{3}{4} + \frac{\sqrt{31} i}{4}$$
$$x_{2} = \frac{3}{4} - \frac{\sqrt{31} i}{4}$$
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

$$2 \cdot 0^{2} - 3 \cdot 0 + 5 \geq 0$$
5 >= 0

so the inequality is always executed
Solving inequality on a graph
Rapid solution 2 [src]
(-oo, oo)
$$x\ in\ \left(-\infty, \infty\right)$$
x in Interval(-oo, oo)
Rapid solution [src]
And(-oo < x, x < oo)
$$-\infty < x \wedge x < \infty$$
(-oo < x)∧(x < oo)
The graph
2x^2-3x+5>=0 inequation