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

Nov 23, 2016

$4 {x}^{2} + 7 x < - 3$ is true is when x is on the interval $- 1 < x < - \frac{3}{4}$

#### Explanation:

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

$4 {x}^{2} + 7 x + 3 < 0$
$4 {x}^{2} + 7 x + 3 = 0$
$4 {x}^{2} + 4 x + 3 x + 3 = 0$
$4 x \left(x + 1\right) + 3 \left(x + 1\right) = 0$
$\left(4 x + 3\right) \left(x + 1\right) = 0$
$x = - 1 , x = - \frac{3}{4}$
$- 1 < x < - \frac{3}{4}$