Помогите решить через дискриминант.

$1)5x + 6 = {x}^{2} \\ 5x + 6 - {x}^{2} = 0 \\ - {x}^{2} + 5x + 6 = 0 \times ( - 1)$
${x}^{2} - 5x - 6 = 0 \\ d = 25 + 24 = 49 = {7}^{2} \\$
$x1 = \frac{5 + 7}{2} = \frac{12}{2} = 6 \\ x2 = \frac{5 - 7}{2} = - \frac{2}{2} = - 1$
2)
$4x + 14(x - 3) = 3x(x - 3) \\ 4x + 14x - 42 = {3x}^{2} - 9x \\ 18x - 42 = 3 {x}^{2} - 9x \\ 18x - 42 - 3 {x}^{2} + 9x = 0$
$- 3 {x}^{2} + 27x - 42 = 0 \div ( - 3) \\ {x}^{2} - 9x + 14 = 0 \\ d = 81 - 56 = 25 = {5}^{2}$
$x1 = \frac{9 + 5}{2} = \frac{14}{2} = 7 \\ x2 = \frac{9 - 5}{2} = \frac{4}{2} = 2$