Помогите пожалуйста. Решите первый уровень

Решите задачу:

$( \sqrt{} 22 - 3x) {}^{2} = 2 {}^{2} \\ 22 - 3x = 4 \\ x = 6$
$( \sqrt{x + 1} ) {}^{2} = 16 \\ x + 1 = 16 \\ x = 15$
$\frac{20}{3x - 13} = \frac{1}{4} \\ - 3x = - 80 - 13 \\ x = 31$
$21 - 4x = x {}^{2} \\ - x {}^{2} - 4x + 21 = 0 \\ d = 16 + 84 = 100 \\ x1 = \frac{4 - 10}{ - 2} = 3 \\ x2 = \frac{4 + 10}{ - 2} = - 7$
$21 - 4x = {x}^{2} \\ - {x}^{2} - 4x + 21 = 0 \\ x1 = 3 \\ x2 = - 7$
$\frac{4}{7} x = - \frac{32}{7} \\ x = \frac{ - 32 \times 7}{7 \times 4} = - 8$
$x = \frac{6x - 5}{x + 12} \\ {x}^{2} + 6x + 5 = 0 \\ d = 36 - 20 = 16 \\ x1 = \frac{ - 6 - 4}{2} = - 5 \\ x2 = \frac{ - 6 + 4}{2} = - 1$
$\frac{5m {}^{30} + 13m {}^{30} }{4m {}^{30} } = \frac{18m {}^{30} }{4m {}^{30} } = 4.5$