(x+1)(6x(в квадрате)-5x+1)=0

x₁ + 1 =0x₁ = - 16x² - 5x + 1 = 0D = b² - 4ac = (-5)² - 4 × 6 × 1 = 25 - 24 = 1x₂ = ( 5 + 1)/12= 0,5x₃ = ( 5 - 1)/12 = 4/12 = 1/3

[latex](x+1)(6x^2-5x+1)=0 \ \ x+1=0 \ x_1=-1 \ \ 6x^2-5x+1=0 \ D=b^2-4ac=1 \ x_2_,_3= frac{-b^+_- sqrt{D} }{2a} \ x_2= frac{1}{3} \ x_3= frac{1}{2} [/latex]