# How do you solve and write the following in interval notation: 2x^2+5x-12 ≤ 0?

Jun 21, 2016

x in [-4;3/2]

#### Explanation:

You must find the zeroes of the trynomial
$2 {x}^{2} + 5 x - 12$

using the formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case you have:

a=1;b=5;c=-12

so that

$x = \frac{- 5 \pm \sqrt{25 - 4 \left(2\right) \left(- 12\right)}}{2 \left(2\right)}$

$x = \frac{- 5 \pm \sqrt{25 + 96}}{4}$

$x = \frac{- 5 \pm \sqrt{121}}{4}$

$x = \frac{- 5 \pm 11}{4}$

$x = - \frac{16}{4} = - 4 \mathmr{and} x = \frac{6}{4} = \frac{3}{2}$

The trynomial is therefore negative in the interval

[-4;3/2]