Question

решите уравнение (3х-2)(х-3)=20

Answer 1

$\ (3x-2)*(x-3)=20 \ \ 3x^2-9x-2x+6-20=0 \ \ 3x^2-11x-14=0 \ \ D=(-11)^2-4*3*(-14)=121+168=289 \ \ sqrt{289} =17 \ \ x_{1}= frac{11-7}{6} = -1 \ \ x_{2}= frac{11+7}{6} =3$

Ответ: $\ x_{1}= frac{11-7}{6} = -1 \ \ x_{2}= frac{11+7}{6} =3$