2х2≤5+9х решить неравенство

$2\textup{x}^2 \leq 5+9\textup{x} \\ 2\textup{x}^2-9\textup{x}-5 \leq 0 \\ 2\textup{x}^2-9\textup{x}-5=0 \\ \textup{D}=81-4*2*(-5)=81+40=121 \\ \\ \textup{x}_1= \frac{9+11}{4} = \frac{20}{4} =5 ~~~~~~~~~~~~~~~~~~~~~~~~ \textup{x}_2= \frac{9-11}{4} = \frac{-2}{4} =-0,5 \\ \\ 2(\textup{x}+0,5)(\textup{x}-5) \leq 0$

_______+_______ $-0,5$ _______-______ $5$ ______+_______

Ответ: $x\in[-0,5;~5]$