Помогите пожалуйста решить

$1. {x}^{2} - 5x + 6 = 0 \\ d = 25 - 24 = 1 \\ x1 = \frac{5 + 1}{2} = \frac{6}{2} = 3$
$x2 = \frac{5 - 1}{2} = \frac{4}{2} = 2$
$2)(2x - 1)(x - 1) = \\ (3x + 4)(x + 7) = {2x}^{2} - 2x - x \\ + 1 = 3 {x}^{2} + 21x + 4x + 28 \\ 2 {x}^{2} - 3x + 1 = 3 {x}^{2} + 25x + 28$
$2 {x}^{2} -3 x + 1 - 3 {x}^{2} - 25x - 28 = 0 \\ - {x}^{2} - 28x - 27 = 0$
Умножим на - 1, получаем
${x}^{2} + 28x + 27 = 0 \\ d =784 - 108 = 676 = 26 \\ x1 = \frac{ - 28 + 26}{2} = - \frac{2}{2} = - 1$
$x2 = \frac{ - 28 - 26}{2} = - \frac{54}{2} = - 27$
$3)12 = x(7 - x) \\ 7x - {x}^{2} = 12 \\ 7x - {x}^{2} - 12 = 0 \\ - {x}^{2} + 7x - 12 = 0$
Умножим на минус 1
${x}^{2} - 7x + 12 = 0 \\ d = 49 - 48 = 1 \\ x1 = \frac{7 + 1}{2} = \frac{8}{2} = 4$
$x2 = \frac{7 - 1}{2} = \frac{6}{2} = 3$
$4)3( {x}^{2} + 4x) = 2x(x + 2) \\ 3 {x}^{2} + 12x = 2 {x}^{2} + 4x \\ 3 {x}^{2} + 12x - 2 {x}^{2} - 4x = 0 \\ {x}^{2} + 8x = 0$
$x(x + 8) = 0 \\ x = 0 \\ x + 8 = 0 \\ x = - 8$
$5)(x - 2)(x - 4) = (x + 3)(x + 2) \\ {x}^{2} - 4x - 2x + 8 = {x}^{2} + 2x + 3x \\ + 6 \\ {x}^{2} - 6x + 8 = {x}^{2} + 5x + 6$
${x}^{2} - 6x + 8 - {x}^{2} - 5x - 6 = 0 \\ - 11x + 2 = 0 \\ - 11x = - 2 \\ x = \frac{2}{11}$