$x^2 + 3x\sqrt{2} + 4 = 0$
$D = b^2 - 4ac = (3\sqrt{2})^2 - 4 * 1 * 4 = 9 * 2 - 16 = 18 - 16 = 2 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3\sqrt{2} + \sqrt{2}}{2 * 1} = \frac{-2\sqrt{2}}{2} = -\sqrt{2}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3\sqrt{2} - \sqrt{2}}{2 * 1} = \frac{-4\sqrt{2}}{2} = -2\sqrt{2}$
Ответ: $x = -2\sqrt{2}$ и $x = -\sqrt{2}$

$x^2 - x(\sqrt{3} + 2) + 2\sqrt{3} = 0$
$D = b^2 - 4ac = (\sqrt{3} + 2)^2 - 4 * 1 * 2\sqrt{3} = 3 + 4\sqrt{3} + 4 - 8\sqrt{3} = 3 - 4\sqrt{3} + 4 = (\sqrt{3} - 2)^2 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{\sqrt{3} + 2 + \sqrt{(\sqrt{3} - 2)^2}}{2 * 1} = \frac{\sqrt{3} + 2 + \sqrt{3} - 2}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{\sqrt{3} + 2 - \sqrt{(\sqrt{3} - 2)^2}}{2 * 1} = \frac{\sqrt{3} + 2 - (\sqrt{3} - 2)}{2} = \frac{\sqrt{3} + 2 - \sqrt{3} + 2}{2} = \frac{4}{2} = 2$
Ответ: $x = \sqrt{3}$ и x = 2

$\frac{2x^2 + x}{3} - \frac{x + 3}{4} = x - 1$ |* 12
$4(2x^2 + x) - 3(x + 3) = 12(x - 1)$
$8x^2 + 4x - 3x - 9 = 12x - 12$
$8x^2 + x - 9 - 12x + 12 = 0$
$8x^2 - 11x + 3 = 0$
$D = b^2 - 4ac = (-11)^2 - 4 * 8 * 3 = 121 - 96 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{11 + \sqrt{25}}{2 * 8} = \frac{11 + 5}{16} = \frac{16}{16} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{11 - \sqrt{25}}{2 * 8} = \frac{11 - 5}{16} = \frac{6}{16} = \frac{3}{8}$
Ответ: $x = \frac{3}{8}$ и x = 1