Answer
$-1 \leq x \leq \frac{1}{2}$
The solution set is $~~[-1, \frac{1}{2}]$
Work Step by Step
$2x^2+x \leq 1$
$2x^2+x-1 \leq 0$
We can use the quadratic formula to find the values of $x$ such that $2x^2+x-1 = 0$:
$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$x = \frac{-1 \pm \sqrt{(1)^2-4(2)(-1)}}{(2)(2)}$
$x = \frac{-1 \pm \sqrt{9}}{4}$
$x = \frac{-1 \pm 3}{4}$
$x = -1, \frac{1}{2}$
When $~~x=-1~~$ or $~~x=\frac{1}{2},~~$ then $~~2x^2+x-1 = 0$
When $x \lt -1,~~$ then $~~2x^2+x-1 \gt 0$
When $-1 \lt x \lt \frac{1}{2},~~$ then $~~2x^2+x-1 \lt 0$
When $x \gt \frac{1}{2},~~$ then $~~2x^2+x-1 \gt 0$
Solution:
$-1 \leq x \leq \frac{1}{2}$
The solution set is $~~[-1, \frac{1}{2}]$
