Answer
$x \leq -\frac{3}{2}~~$ or $~~x \geq 1$
The solution set is $~~(-\infty, -\frac{3}{2}]\cup [1,\infty)$
Work Step by Step
$(2x+3)(x-1) \geq 0$
When $~~x=-\frac{3}{2}~~$ or $~~x=1,~~$ then $~~(2x+3)(x-1) = 0$
When $x \lt -\frac{3}{2},~~$ then $~~(2x+3)(x-1) \gt 0$
When $-\frac{3}{2} \lt x \lt 1,~~$ then $~~(2x+3)(x-1) \lt 0$
When $x \gt 1,~~$ then $~~(2x+3)(x-1) \gt 0$
Solution:
$x \leq -\frac{3}{2}~~$ or $~~x \geq 1$
The solution set is $~~(-\infty, -\frac{3}{2}]\cup [1,\infty)$
