Question

решить
3x^2+4x-4=0
(x-3)(x-2)=6(x-3)

Answer 1

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