2b^2+4b-126=0

解

2 b^{2} + 4 b - 126 = 0

b = 7, b = - 9

解答ステップ

2 b^{2} + 4 b - 126 = 0

解くとthe二次式

b_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 126 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 126) = 32

b_{1, 2} = \frac{- 4 \pm 32}{2 \cdot 2}

解を分離する

b_{1} = \frac{- 4 + 32}{2 \cdot 2}, b_{2} = \frac{- 4 - 32}{2 \cdot 2}

b = \frac{- 4 + 32}{2 \cdot 2} : 7

b = \frac{- 4 - 32}{2 \cdot 2} : - 9

二次equationの解:

b = 7, b = - 9