Решить3x^2+4x-4=0(x-3)(x-2)=6(x-3)

1) $3x^2+4x-4=0$

D = $4^{2}$ + -4 *(-4) * 3= 48 + 16= 64

$\frac{-4±8}{6}$

$\left \{ {{\frac{-4-8}{6}=\frac{-12}{6}=-2 } \atop {\frac{-4+8}{6}=\frac{4}{6}=\frac{2}{3}}} \right.$

Ответ : - 2, $\frac{2}{3}$

2)(x-3)(x-2)=6(x-3)\\x^{2} -2x+6 -3x = 6x - 18\\x^{2} -11x +24=0\\[/tex]

D= $121 - 96= 25$

$\frac{11 ± 5}{2}$

$\frac{11 - 5}{2} =4,5\\\frac{11 + 5}{2} = 8$

Ответ: 4,5, 8