How do you solve quadratic inequalities with imaginary numbers?

Question

Examples:

5x^2 + 3x ≥ 0

5x^2 + 3x ≤ 0

when solving for x when they're equal to zero, we get an answer with i. how do we solve for the solution set then?

Answer 1

Complex numbers do not form an ordered field, so you cannot compare them to each other like that. e.g. 2i is not grater than i

Answer 2

These are not complex.

Put equal to 0 and factorise: 5x^2 + 3x =0

                                          x(5x +3) =0

                                          x=0 or x = -3/5

Now draw a u shaped curve with roots 0 and -3/5

This curve is >= 0 above the x axis so x>=0 and x<= -3/5

This curve is <=0 (second example) under the x axis so -3/5 < x < 0