16t^2-128t+192=0

解

16 t^{2} - 128 t + 192 = 0

t = 6, t = 2

解答ステップ

16 t^{2} - 128 t + 192 = 0

解くとthe二次式

t_{1, 2} = \frac{- ( - 128 ) \pm ( - 128 ) ^{2} - 4 \cdot 16 \cdot 192}{2 \cdot 16}

(- 128)^{2} - 4 \cdot 16 \cdot 192 = 64

t_{1, 2} = \frac{- ( - 128 ) \pm 64}{2 \cdot 16}

解を分離する

t_{1} = \frac{- ( - 128 ) + 64}{2 \cdot 16}, t_{2} = \frac{- ( - 128 ) - 64}{2 \cdot 16}

t = \frac{- ( - 128 ) + 64}{2 \cdot 16} : 6

t = \frac{- ( - 128 ) - 64}{2 \cdot 16} : 2

二次equationの解:

t = 6, t = 2