# How do you solve x^2-4x>=5?

Feb 1, 2018

x^2−4x≥5

x^2−4x-5≥0

x^2+x-5x-5≥0

x(x+1)-5(x+1)≥0

(x-5)(x+1)≥0

:.(x-5)≥0 and (x+1)≥0

$x \ge 5$ and $x \ge - 1$

$x \le 5$ as it will also be greater than -1 then.

-Sahar.