Решите уравнение (3х-2)(х-3)=20

[latex] \\ (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 \\ [/latex] Ответ: [latex]\\ x_{1}= \frac{11-7}{6} = -1 \\ \\ x_{2}= \frac{11+7}{6} =3 \\ [/latex]