[latex]8 x^{2} -19x-29 \geq -2 x^{2} [/latex]

[latex]8x^2-19x-29 \geq -2x^2\\10x^2-19x-29 \geq 0\\y=10x^2-19x-29\\y=0,10x^2-19x-29=0\\D=b^2-4ac=361+1160=1521=39^2,D\ \textgreater \ 0\\x_{1,2}= \frac{-bб \sqrt{D} }{2a} = \frac{19б39}{20} =|{ {{x=2.9} \atop {x=-1}} \right. \\10(x+1)(x-2.9) \geq 0 [/latex]

[latex]8x^2-19x-29 \geq -2x^2 \\ 10x^2-19x-29 \geq 0 \\ \\ 10x^2- 19x-29=0 \\ D=b^2-4ac=19^2-4*10*(-29)=361+1160=1521=39^2 \\ \\ x_{1,2}= \frac{-b^+_- \sqrt{D} }{ 2a} \\ \\ x_1= \frac{-(-19)+39}{2*10} =2.9 \\ \\ x_2= \frac{-(-19)-39}{2*10} =-1 \\ \\ \\ 10x^2-19x-29 \geq 0 \\ \\ +++[-1]---[2.9]+++\ \textgreater \ x \\ \\ x\in (-\infty;-1] \ U \ [2.9;+\infty) \\ \\ OTBET: \ x\in (-\infty;-1] \ U \ [2.9;+\infty)[/latex]