$2x^2 + x\sqrt{5} - 15 = 0$
$D = b^2 - 4ac = (\sqrt{5})^2 - 4 * 2 * (-15) = 5 + 120 = 125 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-\sqrt{5} + \sqrt{125}}{2 * 2} = \frac{-\sqrt{5} + \sqrt{25 * 5}}{4} = \frac{-\sqrt{5} + 5\sqrt{5}}{4} = \frac{4\sqrt{5}}{4} = \sqrt{5}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-\sqrt{5} - \sqrt{125}}{2 * 2} = \frac{-\sqrt{5} - \sqrt{25 * 5}}{4} = \frac{-\sqrt{5} - 5\sqrt{5}}{4} = \frac{-6\sqrt{5}}{4} = -\frac{3\sqrt{5}}{2}$
Ответ: $x = -\frac{3\sqrt{5}}{2}$ и $x = \sqrt{5}$

$x^2 - x(\sqrt{6} - 1) - \sqrt{6} = 0$
$D = b^2 - 4ac = (\sqrt{6} - 1)^2 - 4 * 1 * (-\sqrt{6}) = 6 - 2\sqrt{6} + 1 + 4\sqrt{6} = 6 + 2\sqrt{6} + 1 = (\sqrt{6} + 1)^2 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{\sqrt{6} - 1 + \sqrt{(\sqrt{6} + 1)^2}}{2 * 1} = \frac{\sqrt{6} - 1 + \sqrt{6} + 1}{2} = \frac{2\sqrt{6}}{2} = \sqrt{6}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{\sqrt{6} - 1 - \sqrt{(\sqrt{6} + 1)^2}}{2 * 1} = \frac{\sqrt{6} - 1 - \sqrt{6} - 1}{2} = \frac{-2}{2} = -1$
Ответ: x = −1 и $x = \sqrt{6}$

$\frac{x^2 - 4}{8} - \frac{2x + 3}{3} = -1$ |* 24
$3(x^2 - 4) - 8(2x + 3) = -24$
$3x^2 - 12 - 16x - 24 + 24 = 0$
$3x^2 - 16x - 12 = 0$
$D = b^2 - 4ac = (-16)^2 - 4 * 3 * (-12) = 256 + 144 = 400 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{16 + \sqrt{400}}{2 * 3} = \frac{16 + 20}{6} = \frac{36}{6} = 6$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{16 - \sqrt{400}}{2 * 3} = \frac{16 - 20}{6} = \frac{-4}{6} = -\frac{2}{3}$
Ответ: $x = -\frac{2}{3}$ и x = 6

$\frac{4x^2 + x}{3} - \frac{x^2 + 17}{9} = \frac{5x - 1}{6}$ |* 18
$6(4x^2 + x) - 2(x^2 + 17) = 3(5x - 1)$
$24x^2 + 6x - 2x^2 - 34 = 15x - 3$
$22x^2 + 6x - 34 - 15x + 3 = 0$
$22x^2 - 9x - 31 = 0$
$D = b^2 - 4ac = (-9)^2 - 4 * 22 * (-31) = 81 + 2728 = 2809 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{9 + \sqrt{2809}}{2 * 22} = \frac{9 + 53}{44} = \frac{62}{44} = \frac{31}{22} = 1\frac{9}{22}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{9 - \sqrt{2809}}{2 * 22} = \frac{9 - 53}{44} = \frac{-44}{44} = -1$
Ответ: x = −1 и $x = 1\frac{9}{22}$