Question

${x}^{2} + 6x - 12 = 0$
Помогите

Answer 1

Решить квадратное уравнение.

$x^2 + 6x - 12 = 0; \longleftrightarrow \left[ax^2 + bx + c = 0\right]\\D = \left[b^2 - 4ac\right] = 6^2 - 4\cdot(-12) = 36 + 4\cdot12 = 36 + 48 = 84;\\x_{1,2} = \left[\dfrac{-b\pm\sqrt{D}}{2a}\right] = \dfrac{-6\pm2\sqrt{21}}{2} = -3\pm\sqrt{21}.\\\\\left[\begin{array}{c}x_1 = -3 + \sqrt{21},&x_2 = -3 - \sqrt{21}.\end{array}$

$\bf OTBET: -3 + \sqrt{21}; -3 - \sqrt{21}.$