Question

2x^2-7X>15
Let me just break it down till the end.

(x-5)(2x+3)>0

x-5-0 2x+3=0
x=5 x=-3/2

When I do my test point and let it be 0 It looks something like

<--//////-----------------------------------------------------///////--------->
True -3 over 2 the middle is F and 5 is also true and the interval notation is (-∞.-3/2)U(5,∞).

My question is, how can -3/2 be greater than 0?

Go back to (x-5)(2x+3)>0
This is true if both brackets are positive or both brackets negative.
Both positive if x>5 and both negative if x<-3/2

Actually :