Answer
$$\bf\color{blue}{[-\frac{3}{2},6]}$$
Work Step by Step
$\bf\text{First, rewrite the inequality so all terms are on the left side}$
$2x^2-9x-18\leq{0}$
$\bf\text{Now we apply the quadratic formula where:}$
$a=2$, $b=-9$, and $c=-18$
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{9\pm\sqrt{(-9)^2-4(2)(-18)}}{2(2)}$
$x=\dfrac{9\pm\sqrt{81+144}}{4}$
$x=\dfrac{9\pm\sqrt{255}}{4}$
$x=\dfrac{9\pm15}{4}$
$x=\dfrac{24}{4}$ or $x=\dfrac{-6}{4}$
$x=6$ or $x=-\dfrac{3}{2}$
$\bf\text{These values give us three intervals:}$
$(-\infty,-\frac{3}{2}]$ $[-\frac{3}{2},6]$ $[6,\infty)$
$\bf\text{Using test values we can test each interval in the formula}$
For the interval $\bf{(-\infty,-\frac{3}{2}]}$, we can use $-2$ as our test value
$2(-2)^2-9(-2)-18\leq{0}$
$8+18-18\leq{0}$
$8\leq{0}$ which is untrue so this interval is $\bf\text{not valid}$
For the interval $\bf{[-\frac{3}{2},6]}$ we can use $1$ as our test value
$2(1)^2-9(1)-18\leq{0}$
$2-9-18\leq{0}$
$-25\leq{0}$ which is true so this interval is $\bf\text{valid}$
For the interval $\bf{[6,\infty)}$, we can use $7$ as our test value
$2(7)^2-9(7)-18\leq{0}$
$2(49)-9(7)-18\leq{0}$
$98-63-18\leq{0}$
$17\leq{0}$ which is untrue so this interval is $\bf\text{not valid}$
Therefore the solution is $\color{blue}{[-\dfrac{3}{2},6]}$
