Answer
$\color{blue}{\bf{[-2,\dfrac{3}{2}]\cup[3,\infty)}}$
Work Step by Step
$(2x-3)(x+2)(x-3)\geq0$
$\bf\text{Rewrite as an equation and find the zeros.}$
$(2x-3)(x+2)(x-3)=0$
$2x-3=0$
$2x=3$
$\bf{x=\frac{3}{2}}$
$x+2=0$
$\bf{x=-2}$
$x-3=0$
$\bf{x=3}$
$\bf\text{So, we now have the values:}$ {${-2,\frac{3}{2},3}$}
$\bf\text{Which gives us four intervals:}$
$\bf(-\infty,-2]$ $\bf{[-2,\dfrac{3}{2}]}$ $\bf{[\dfrac{3}{2},3]}$ $\bf{[3,\infty]}$
$\bf\text{Now we select test values for each interval}$
which is untrue so this interval is $\bf\text{not valid}$
which is true so this interval is $\bf\text{valid}$
For the interval $\bf(-\infty,-2]$ we can use $\bf{-3}$ as our test value
$(2(-3)-3)(-3+2)(-3-3)\geq0$
$(-6-3)(-3+2)(-3-3)\geq0$
$(-9)(-1)(-6)\geq0$
$-54\geq0$ which is not true, so this interval does not satisfy the inequality
For the interval $\bf{[-2,\dfrac{3}{2}]}$ we can use $\bf{0}$ as our test value
$(2(0)-3)(0+2)(0-3)\geq0$
$(-3)(2)(-3)\geq0$
$18\geq0$ which is true, so this interval satisfies the inequality
For the interval $\bf{[\dfrac{3}{2},3]}$ we can use $\bf{2}$ as our test value
$(2(2)-3)(2+2)(2-3)\geq0$
$(4-3)(2+2)(2-3)\geq0$
$(1)(4)(-1)\geq0$
$-4\geq0$ which is not true, so this interval does not satisfy the inequality
For the interval $\bf{[3,\infty]}$ we can use $\bf{4}$ as our test value
$(2(4)-3)(4+2)(4-3)\geq0$
$(8-3)(4+2)(4-3)\geq0$
$(5)(6)(1)\geq0$
$30\geq0$ which is true, so this interval satisfies the inequality
Therefore, $\color{blue}{\bf{[-2,\dfrac{3}{2}]\cup[3,\infty)}}$ satisfies the inequality
