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

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Answer 1 · 2016-11-23T01:37:58

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

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

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.

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

How do you solve 4x^2+7x<-34x^2+7x<-3?