Вопрос в картинках...

$( x - 2)(x^2 + 6x + 9 ) - 6( x + 3) = 0$
$( x - 2)( x + 3)^2 - 6( x + 3) = 0$
$( x + 3)( (x-2)(x+3) - 6) = 0$
$( x + 3)( x^2 + x - 6 - 6) = 0$
$( x + 3)( x^2 + x - 12) = 0$
$x + 3 = 0$
$x = - 3$
$x^2 + x - 12 = 0$
$D = b^2 - 4ac = 1 + 48 = 49 = 7^2$
$x_{1} = ( - 1 + 7) / 2 = 3$
$x_{2} = ( - 1 - 7) / 2 = - 4$
Ответы: $x_{1} = 3///x_{2} = - 4///x_{3} = - 3.$