Question

How do you solve `-s^2+4s-6>0`?

Answer 1

Never

Explanation:

Using the discriminant to determine how many zeros the function has.

`D=sqrt(b^2-4ac)`

where `b=4`, `a=-1` and `c=-6`

`D=sqrt((4)^2-4(-1)(-6))`

`D=sqrt(16+(-24)`

`D=sqrt-8`

Since the discriminant is negative, there are no real roots for this function (the quadratic will never cross the x-axis). Therefore the function will NEVER be greater than `0`

Here is a graphical look,

graph{-x^2+4x-6 [-10, 10, -5, 5]}