-5t^2+80t=0

解

- 5 \cdot t^{2} + 80 \cdot t = 0

t = 0, t = 16

解答ステップ

- 5 t^{2} + 80 t = 0

解くとthe二次式

t_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 ( - 5 ) \cdot 0}{2 ( - 5 )}

8 0^{2} - 4 (- 5) \cdot 0 = 80

t_{1, 2} = \frac{- 80 \pm 80}{2 ( - 5 )}

解を分離する

t_{1} = \frac{- 80 + 80}{2 ( - 5 )}, t_{2} = \frac{- 80 - 80}{2 ( - 5 )}

t = \frac{- 80 + 80}{2 ( - 5 )} : 0

t = \frac{- 80 - 80}{2 ( - 5 )} : 16

二次equationの解:

t = 0, t = 16