16t^2-80t+64=0

解

16 t^{2} - 80 t + 64 = 0

t = 4, t = 1

解答ステップ

16 t^{2} - 80 t + 64 = 0

解くとthe二次式

t_{1, 2} = \frac{- ( - 80 ) \pm ( - 80 ) ^{2} - 4 \cdot 16 \cdot 64}{2 \cdot 16}

(- 80)^{2} - 4 \cdot 16 \cdot 64 = 48

t_{1, 2} = \frac{- ( - 80 ) \pm 48}{2 \cdot 16}

解を分離する

t_{1} = \frac{- ( - 80 ) + 48}{2 \cdot 16}, t_{2} = \frac{- ( - 80 ) - 48}{2 \cdot 16}

t = \frac{- ( - 80 ) + 48}{2 \cdot 16} : 4

t = \frac{- ( - 80 ) - 48}{2 \cdot 16} : 1

二次equationの解:

t = 4, t = 1