Пожжжжжжалуйста )))))

[latex]3x^2-2x-5\ \textgreater \ 0 \\ D=(-2)^2+5*3*4=4+60=64 \\ x_{1,2} = \frac{2б8}{6} \\ x_1 = 10/6 ; x_2 = -1 \\ x $\in$ (-\infty;-1) \cup (\frac{10}{6};+\infty) \\ \\ x^2+6x+9\ \textless \ 0 \\ D=6^2-4*9=0 \\ x = \frac{-6}{2} = -3 \\ (x+3)^2\ \textless \ 0 \\ x $\in$ \emptyset \\ \\ -x^2+6x \geq 0 \\ x(-x+6) \geq 0 \\ x=0;x=6 \\ x $\in$ [0;6] [/latex]