Question

Докажите тождество  10x2+19x-2=10(x-0.1)(x+2)

Answer 1

$10x^2+19x-2=10(x-0.1)(x+2)\ 10x^2+19x-2=0\ D=19^2-4*10*(-2)=361+80=441\ x_1= frac{-19-21}{2*10}= -frac{40}{20}=-2\ x_2= frac{-19+21}{2*10}= frac{2}{20}=0.1\ 10x^2+19x-2=0=10(x+2)(x-0.1)$