(10-x)(-12-x)+40=0

解

(10 - x) (- 12 - x) + 40 = 0

x = 8, x = - 10

解答ステップ

(10 - x) (- 12 - x) + 40 = 0

拡張

(10 - x) (- 12 - x) + 40 : x^{2} + 2 x - 80

x^{2} + 2 x - 80 = 0

解くとthe二次式

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

2^{2} - 4 \cdot 1 \cdot (- 80) = 18

x_{1, 2} = \frac{- 2 \pm 18}{2 \cdot 1}

解を分離する

x_{1} = \frac{- 2 + 18}{2 \cdot 1}, x_{2} = \frac{- 2 - 18}{2 \cdot 1}

x = \frac{- 2 + 18}{2 \cdot 1} : 8

x = \frac{- 2 - 18}{2 \cdot 1} : - 10

二次equationの解:

x = 8, x = - 10