Answer
$[2, 5]$
Work Step by Step
$x^2-7x+10\leq0$
$(x-2)(x-5) \leq 0$
$x-2=0$
$x=2$
$x-5=0$
$x=5$
(-infinity, $2]$
$[2, 5]$
$[5$, infinity)
Let $x=0$, $x=3$, and $x=10$
$x=0$
$(x-2)(x-5) \leq 0$
$(0-2)(0-5) \leq 0$
$-2*-5\leq 0$
$10 \leq 0$ (false)
$x=3$
$(x-2)(x-5) \leq 0$
$(3-2)(3-5) \leq 0$
$1*-2\leq 0$
$-2 \leq 0$ (true)
$x=10$
$(x-2)(x-5) \leq 0$
$(10-2)(10-5) \leq 0$
$8*5\leq 0$
$40 \leq 0$ (false)
