t^2-12t+27=0

解

t^{2} - 12 t + 27 = 0

t = 9, t = 3

解答ステップ

t^{2} - 12 t + 27 = 0

解くとthe二次式

t_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 27}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 27 = 6

t_{1, 2} = \frac{- ( - 12 ) \pm 6}{2 \cdot 1}

解を分離する

t_{1} = \frac{- ( - 12 ) + 6}{2 \cdot 1}, t_{2} = \frac{- ( - 12 ) - 6}{2 \cdot 1}

t = \frac{- ( - 12 ) + 6}{2 \cdot 1} : 9

t = \frac{- ( - 12 ) - 6}{2 \cdot 1} : 3

二次equationの解:

t = 9, t = 3