x^2+2x(-x+2)-3(-x+2)^2=0

解

x^{2} + 2 x (- x + 2) - 3 (- x + 2)^{2} = 0

x = 1, x = 3

解答ステップ

x^{2} + 2 x (- x + 2) - 3 (- x + 2)^{2} = 0

拡張

x^{2} + 2 x (- x + 2) - 3 (- x + 2)^{2} : - 4 x^{2} + 16 x - 12

- 4 x^{2} + 16 x - 12 = 0

解くとthe二次式

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 4 ) ( - 12 )}{2 ( - 4 )}

1 6^{2} - 4 (- 4) (- 12) = 8

x_{1, 2} = \frac{- 16 \pm 8}{2 ( - 4 )}

解を分離する

x_{1} = \frac{- 16 + 8}{2 ( - 4 )}, x_{2} = \frac{- 16 - 8}{2 ( - 4 )}

x = \frac{- 16 + 8}{2 ( - 4 )} : 1

x = \frac{- 16 - 8}{2 ( - 4 )} : 3

二次equationの解:

x = 1, x = 3