Answer
$$\color{blue}{[-2,3]}$$
Work Step by Step
$\bf\text{First, simplify and rewrite with all terms on the left side.}$
$x(x-1)\leq6$
$x^2-x\leq6$
$x^2-x-6\leq0$
$\bf\text{Factoring the trinomial gives us:}$
$x=-2$ or $x=3$
$\bf\text{Which gives us three intervals:}$
$\bf(-\infty,-2]$ $\bf[-2,3]$ $\bf[3,\infty)$
$\bf\text{Using test values we can test each interval in the formula}$
For the interval $\bf(-\infty,-2]$, we can use $-3$ as our test value
$(-3)^2-(-3)-6\leq0$
$9+3-6\leq0$
$6\leq0$ which is untrue so this interval is $\bf\text{not valid}$
For the interval $\bf[-2,3]$ we can use $1$ as our test value
$(1)^2-(1)-6\leq0$
$1-7\leq0$
$-6\leq0$ which is true so this interval is $\bf\text{valid}$
For the interval $\bf[3,\infty)$, we can use $4$ as our test value
$(4)^2-(4)-6\leq0$
$16-10\leq0$
$6\leq0$ which is untrue so this interval is $\bf\text{not valid}$
Therefore the solution is $\color{blue}{[-2,3]}$
