# How do you solve 15-2y-y^2<=0?

Dec 22, 2016

The answer is y in ] -oo,-5 ] uu [3, +oo[

#### Explanation:

Let's factorise the equation

$15 - 2 y - {y}^{2} = \left(5 + y\right) \left(3 - y\right) \le 0$

Let $f \left(y\right) = \left(5 + y\right) \left(3 - y\right)$

Now, we can make a sign chart

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 5$$\textcolor{w h i t e}{a a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$5 + y$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$3 - y$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(y\right)$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a}$color(white)(aaa)-

Therefore,

$f \left(y\right) \le 0$, when y in ] -oo,-5 ] uu [3, +oo[

graph{(5+x)(3-x) [-32.47, 32.48, -16.25, 16.24]}