Question

Рівняння 6х(квадрат)+х-7=0

Answer 1

$\boldsymbol{6x^2+x-7 = 0.} \\ \\ \\ D = b^2 - 4ac. \\ D = 1^2 - 4 \cdot 6 \cdot (-7) = 1 + 168 = 169 = 13^2. \\ D \ \textgreater \ 0. \\ \\ \\ x_1_,_2 = \dfrac{-b б \sqrt{D}}{2a}. \\ \\ \\ x_1 = \dfrac{-1+13}{2\cdot6} = \dfrac{-12}{12} = -1. \\ \\ x_2 = \dfrac{-1-13}{2 \cdot 6} = \dfrac{-14}{12} = -\dfrac{7}{6}. \\ \\ \\ \boldsymbol{OTBET: - \dfrac{7}{6}; -1}$

Answer 2

6х² + x - 7 = 0
D = 1 + 168 = 169
x1 = (-1 + 13)/12 = 1
x2 = (-1 - 13)/12 = -14/12 = -7/6
Ответ: -7/6 ; 1