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

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Answer 1 · 2017-07-30T12:45:39

Solution: $x in RR$ or $(-oo,oo)$

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

Vertex $(x) = -b/2a= -4/-2= 2$ .

Vertex $(y) = - 2^2+4*-2 -6= -4+8-6 =-2$

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 $x in RR$ or $(-oo,oo)$ . The graph also shows

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

Solution: $x in RR$ or $(-oo,oo)$

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

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