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Решите задачу:

$480.) \\ \frac{3}{x} + \frac{3}{x + 2} = 4 \\ \frac{3}{x} + \frac{3}{x + 2} - 4 = 0 \\ \frac{3(x + 2) + 3x - 4x(x + 2)}{x(x + 2)} = 0 \\ \frac{3x + 6 + 3x - 4x {}^{2} - 8x}{x(x +2)} = 0 \\ \frac{ - 2x + 6 - 4x {}^{2} }{x(x + 2)} = 0 \\ - 2x + 6 - 4x {}^{2} = 0 \\ - 4x {}^{2} - 2x + 6 = 0 \\ 2x {}^{2} + x - 3 = 0 \\ d = 1 {}^{2} + 4 \times 2 \times 3 = 1 + 24 = 25 \\ \sqrt{25} = 5 \\ x1 = \frac{ - 1 - 5}{4} = \frac{ - 6}{4} = \frac{ - 3}{2} = - 1.5 \\ x2 = \frac{ - 1 + 5}{4} = \frac{4}{4} = 1 \\ 493.) \\ \frac{2x {}^{2} + 9x }{x - 3} = 0 \\ 2x {}^{2} + 9x = 0 \\ x(2x + 9) = 0 \\ x = 0 \\ 2x + 9 = 0 \\ x = - \frac{9}{2} = - 4.5$