Answer
$[-5/4, 3/2]$
Work Step by Step
$(2x-3)(4x+5) \le 0$
$2x-3 \le 0$
$2x-3+3 \le 0+3$
$2x \le 3$
$2x/2 \le3/2$
$x \le 3/2$
$4x+5\le 0$
$4x+5-5\le 0-5$
$4x \le -5$
$4x/4 \le -5/4$
$x \le -5/4$
Thus, we have three sections: $(-∞, -5/4]$, $[-5/4, 3/2]$, and $[3/2,∞)$. We need to test one value for $x$ in each section to determine if the section would be a solution set. Since we have the $\le$ sign, we include the end points and use brackets instead of parentheses.
Let $x=-2$, $x=0$, and $x=2$
$x=-2$
$(2x-3)(4x+5) \le 0$
$(2*-2-3)(4*-2+5) \le 0$
$(-4-3)(-8+5) \le 0$
$-7*-3 \le 0$
$21 \le 0$ (false)
$x=0$
$(2x-3)(4x+5) \le 0$
$(2*0-3)(4*0+5) \le 0$
$(0-3)(0+5) \le 0$
$-3*5 \ le 0$
$-15\le 0$ (true)
$x=2$
$(2x-3)(4x+5) \le 0$
$(2*2-3)(4*2+5) \le 0$
$(4-3)(8+5)\le0$
$1*13 \le 0$
$13 \le 0$ (false)
