(2x-5)(x+2)-(x+2)(3x-1)=0

解

(2 x - 5) (x + 2) - (x + 2) (3 x - 1) = 0

x = - 4, x = - 2

解答ステップ

(2 x - 5) (x + 2) - (x + 2) (3 x - 1) = 0

拡張

(2 x - 5) (x + 2) - (x + 2) (3 x - 1) : - x^{2} - 6 x - 8

- x^{2} - 6 x - 8 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 1 ) ( - 8 )}{2 ( - 1 )}

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

x_{1, 2} = \frac{- ( - 6 ) \pm 2}{2 ( - 1 )}

解を分離する

x_{1} = \frac{- ( - 6 ) + 2}{2 ( - 1 )}, x_{2} = \frac{- ( - 6 ) - 2}{2 ( - 1 )}

x = \frac{- ( - 6 ) + 2}{2 ( - 1 )} : - 4

x = \frac{- ( - 6 ) - 2}{2 ( - 1 )} : - 2

二次equationの解:

x = - 4, x = - 2