Question

${x}^{2} + 6x - 12 = 0$
Помогите

Answer 1

Решить квадратное уравнение.

$x^2 + 6x - 12 = 0; longleftrightarrow left[ax^2 + bx + c = 0right]\D = left[b^2 - 4acright] = 6^2 - 4cdot(-12) = 36 + 4cdot12 = 36 + 48 = 84;\x_{1,2} = left[dfrac{-bpmsqrt{D}}{2a}right] = dfrac{-6pm2sqrt{21}}{2} = -3pmsqrt{21}.\\left[begin{array}{c}x_1 = -3 + sqrt{21},&x_2 = -3 - sqrt{21}.end{array}$

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