# How do you solve -x^2-2x+8<=0 by graphing?

Write this as follows

$- {x}^{2} - 2 x + 8 \le 0$

${x}^{2} + 2 x + 1 - 1 - 8 \le 0$

${\left(x + 1\right)}^{2} - {3}^{2} \le 0$

$\left(x + 1 + 3\right) \cdot \left(x + 1 - 3\right) \le 0$

$\left(x + 4\right) \cdot \left(x - 2\right) \le 0$

Hence the solutions for the inequality are

$x \ge 2$ and $x \le - 4$

A graphical representation can be seen on the figure below