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

Jul 30, 2017

Solution: $x \in \mathbb{R}$ or $\left(- \infty , \infty\right)$

Explanation:

$- {s}^{2} + 4 s - 6 < 0 a = - 1 , b = 4 , c = - 6$

Vertex $\left(x\right) = - \frac{b}{2} a = - \frac{4}{-} 2 = 2$ .

Vertex $\left(y\right) = - {2}^{2} + 4 \cdot - 2 - 6 = - 4 + 8 - 6 = - 2$

Vertex is $2 , - 2$ . The parabola opens down wards since $a < 0$

The range is $< - 2$ i.e $< 0$ . The domain is any real value

i.e $x \in \mathbb{R}$ or $\left(- \infty , \infty\right)$ . The graph also shows

$- {s}^{2} + 4 s - 6$ is $< - 2$ i.e $< 0$ .

Solution: $x \in \mathbb{R}$ or $\left(- \infty , \infty\right)$

graph{-x^2+4x-6 [-12.66, 12.65, -6.33, 6.33]}