х^(2)-0,5x-5<0 x2 - 0.5x - 5 = 0 D = b2 - 4ac D = 0.25 + 20 = 20.25 = (√20.25)^2 x1,2 = -b ± √D/2a x1 = 0.5 + √20.25/2 x2 = 0.5 - √20.25/2 Ответ: x1 = 0.5 + √20.25/2 ; x2 = 0.5 - √20.25/2 x2 - 0.5x - 5 =a(x-x1)(x+x2)=(x-0.5 + √20.25/2)(x+0.5 - √20.25/2) (x-0.5 + √20.25/2)(x+0.5 - √20.25/2)<0