# How do you solve -p^2+10p-7<=14?

Jan 13, 2017

The answer is p in ] -oo,3 ]uu [7,+ oo[

#### Explanation:

Let's rewrite the equation

$- {p}^{2} + 10 p - 7 \le 14$, $\Leftrightarrow$, ${p}^{2} - 10 p + 21 \ge 0$

Let 's factorise the LHS

$\left(p - 3\right) \left(p - 7\right) \ge 0$

Let $f \left(p\right) = \left(p - 3\right) \left(p - 7\right)$

Now we can make the sign chart

$\textcolor{w h i t e}{a a a a}$$p$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a a}$$7$$\textcolor{w h i t e}{a a a a}$$+ \infty$

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

$\textcolor{w h i t e}{a a a a}$$p - 7$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(p\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(p\right) \ge$, when  p in ] -oo,3 ]uu [7,+ oo[