Question

(x+1)(6x(в квадрате)-5x+1)=0

Answer 1

X₁ + 1 =0
x₁ = - 1
6x² - 5x + 1 = 0
D = b² - 4ac = (-5)² - 4 × 6 × 1 = 25 - 24 = 1
x₂ = ( 5 + 1)/12= 0,5
x₃ = ( 5 - 1)/12 = 4/12 = 1/3

Answer 2

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