# How do you solve 6x^2-5x-4<=0?

May 5, 2015

It is fairly easy to factor the left side of this inequality
$6 {x}^{2} - 5 x - 4 \le 0$
becomes
$\left(3 x - 4\right) \left(2 x + 1\right) \le 0$

The trick is in seeing what this means.

For this to be true
either $\left(3 x - 4\right) \le 0 \rightarrow x \le \frac{4}{3}$
or $\left(2 x + 1\right) \le 0 \rightarrow x < - \frac{1}{2}$
BUT NOT both $< 0$