4u^2+12u=0

解

4 u^{2} + 12 u = 0

u = 0, u = - 3

解答ステップ

4 u^{2} + 12 u = 0

解くとthe二次式

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

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

u_{1, 2} = \frac{- 12 \pm 12}{2 \cdot 4}

解を分離する

u_{1} = \frac{- 12 + 12}{2 \cdot 4}, u_{2} = \frac{- 12 - 12}{2 \cdot 4}

u = \frac{- 12 + 12}{2 \cdot 4} : 0

u = \frac{- 12 - 12}{2 \cdot 4} : - 3

二次equationの解:

u = 0, u = - 3