# How do you solve the inequality x^2+3x-28<0?

Apr 4, 2018

$- 7 < x < + 4$

#### Explanation:

${x}^{2} + 3 x - 28 = {x}^{2} + 7 x - 4 x - 48 = x \left(x + 7\right) - 4 \left(x + 7\right) = \left(x - 4\right) \left(x + 7\right)$

So, the inequality is

$\left(x - 4\right) \left(x + 7\right) < 0$

Now. the left hand side vanishes for $x = - 7$ and $x = + 4$.

If $x < - 7$, then both the factors $\left(x + 7\right)$ and $\left(x - 4\right)$ are negative, and the product is positive. On the other hand, $\mathmr{if} x > + 4$, both factors are positive, leading once again to a positive product.

Thus the inequality is only satisfied when

$- 7 < x < + 4$

where the facor $\left(x + 7\right)$ is positive and $\left(x - 4\right)$ is negative.

You can see this in the graph below

graph{x^2+3x-28 [-10, 10, -40, 40]}