-16t^2+16t+320=0

解

- 16 t^{2} + 16 t + 320 = 0

t = - 4, t = 5

解答ステップ

- 16 t^{2} + 16 t + 320 = 0

解くとthe二次式

t_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 16 ) \cdot 320}{2 ( - 16 )}

1 6^{2} - 4 (- 16) \cdot 320 = 144

t_{1, 2} = \frac{- 16 \pm 144}{2 ( - 16 )}

解を分離する

t_{1} = \frac{- 16 + 144}{2 ( - 16 )}, t_{2} = \frac{- 16 - 144}{2 ( - 16 )}

t = \frac{- 16 + 144}{2 ( - 16 )} : - 4

t = \frac{- 16 - 144}{2 ( - 16 )} : 5

二次equationの解:

t = - 4, t = 5