What is the solution set of the inequality -2x^2 +3x+ 5 > 0?

(-2x + 5)(x + 1) = 0

X = 5/2 or x = -1

So, you check points in the ranges (-infinity, -1), (-1, 5/2), (5/2, infinity)...

X=-2: false

X=0: true

X=3: false

So, -1<x<5/2

There is a very well known rule concerning quadratic functions : The sign of f(x) is the same as the sign of x^2 except between the real roots of the equation f(x) = 0; what that means in this case is that the sign of x^2 is -, the whole function is -ve except between the points where the graph crosses the x-axis. The graph crosses the x-axis where -2x^2 + 3x + 5= 0 or if you prefer, where 2x^2 - 3x - 5 = 0

This can be factorised to (2x - 5)(x + 1) = 0 so the zeros of the function are -1 and 5/2 i.e. the graph crosses the x-axis at x = -1 and at x = 5/2 so the graph is only above the x-axis for -1 < x < 5/2.

or the Solution Set is {x:x in R, -1< x < 5/2} - unfortunately I can't type the symbol for "is a member of".

-1 < x < 5/2

(-1, 5/2)