Question

How do you solve `4x^2+7x<-3`?

Answer 1

`4x^2+7x<-3` is true is when x is on the interval `-1< x < -3/4`

Explanation:

`4x^2+7x<-3`

Add 3 to each side:
`4x^2+7x+3<0`

Logically, this is an upward facing parabola, so to check where the parabola is less than zero, it is the interval between the zeros of the parabola.

`4x^2+7x+3=0`
`4x^2+4x+3x+3=0`
`4x(x+1)+3(x+1)=0`
`(4x+3)(x+1)=0`
`x=-1, x=-3/4`
`-1< x < -3/4`