# How do you solve -2x^2+10<8x using a sign chart?

Oct 29, 2016

The solution are $x < - 5$ and $x > 1$

#### Explanation:

Let's rewrite the expression
$f \left(x\right) = 2 {x}^{2} + 8 x - 10 > 0$
so factorising
$\left(2 x - 2\right) \left(x + 5\right) = 2 \left(x - 1\right) \left(x + 5\right) > 0$
So the values of x to be taken in consideration are $x = 1$ and $x = - 5$

Let 's do the sign chart
$x$$\textcolor{w h i t e}{a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 5$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$x - 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
$x + 5$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

So the values of x are $x < - 5$ and $x > 1$