t^2+20t=0

解

t^{2} + 20 t = 0

t = 0, t = - 20

解答ステップ

t^{2} + 20 t = 0

解くとthe二次式

t_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot 0 = 20

t_{1, 2} = \frac{- 20 \pm 20}{2 \cdot 1}

解を分離する

t_{1} = \frac{- 20 + 20}{2 \cdot 1}, t_{2} = \frac{- 20 - 20}{2 \cdot 1}

t = \frac{- 20 + 20}{2 \cdot 1} : 0

t = \frac{- 20 - 20}{2 \cdot 1} : - 20

二次equationの解:

t = 0, t = - 20