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

Apr 10, 2017

Never

#### Explanation:

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

$D = \sqrt{{b}^{2} - 4 a c}$

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

$D = \sqrt{{\left(4\right)}^{2} - 4 \left(- 1\right) \left(- 6\right)}$

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]}