Question

$8 x^{2} -19x-29 geq -2 x^{2}$

Answer 1

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

Answer 2

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