(t+4)^2-12(t+4)+35=0

解

(t + 4)^{2} - 12 (t + 4) + 35 = 0

t = 3, t = 1

解答ステップ

(t + 4)^{2} - 12 (t + 4) + 35 = 0

拡張

(t + 4)^{2} - 12 (t + 4) + 35 : t^{2} - 4 t + 3

t^{2} - 4 t + 3 = 0

解くとthe二次式

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 3 = 2

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

解を分離する

t_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 1}, t_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 1}

t = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

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

二次equationの解:

t = 3, t = 1