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

1 Answer
Apr 10, 2017

Answer:

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]}