2y^2+12y=0

解

2 y^{2} + 12 y = 0

y = 0, y = - 6

解答ステップ

2 y^{2} + 12 y = 0

解くとthe二次式

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 \cdot 0 = 12

y_{1, 2} = \frac{- 12 \pm 12}{2 \cdot 2}

解を分離する

y_{1} = \frac{- 12 + 12}{2 \cdot 2}, y_{2} = \frac{- 12 - 12}{2 \cdot 2}

y = \frac{- 12 + 12}{2 \cdot 2} : 0

y = \frac{- 12 - 12}{2 \cdot 2} : - 6

二次equationの解:

y = 0, y = - 6