# How do you solve -x^2+x+5>0 by graphing?

Nov 21, 2016

-1.791 < x < 2.791
graph{-x^2+x +5 [-10, 10, -5, 5]}

#### Explanation:

Find the points of intersection of the graph with the x-axis.

$- {x}^{2} + x + 5 = 0$

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

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(- 1\right) 5}}{- 2 \left(- 1\right)}$

$x = \frac{- 1 \pm \sqrt{21}}{2}$

$x = - 1.791 , 2.791$

The point of intersection of the graph with y-axis
is $y = 5$ by substituting $x = 0$ into $- {x}^{2} + x + 5$

Since the coefficient of ${x}^{2}$ is negative, the curve upwards, as shown in the graph.

For $- {x}^{2} + x + 5 > 0$, we look at the region of the graph where the curve is above the y-axis.

Hence, -1.791 < x < 2.791