t^2+2t-8=0

解

t^{2} + 2 t - 8 = 0

t = 2, t = - 4

解答ステップ

t^{2} + 2 t - 8 = 0

解くとthe二次式

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

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

t_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 1}

解を分離する

t_{1} = \frac{- 2 + 6}{2 \cdot 1}, t_{2} = \frac{- 2 - 6}{2 \cdot 1}

t = \frac{- 2 + 6}{2 \cdot 1} : 2

t = \frac{- 2 - 6}{2 \cdot 1} : - 4

二次equationの解:

t = 2, t = - 4