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

解

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

x = 3, x = - 2

解答ステップ

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

拡張

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

x^{2} - x - 6 = 0

解くとthe二次式

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 6) = 5

x_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

解を分離する

x_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

x = \frac{- ( - 1 ) + 5}{2 \cdot 1} : 3

x = \frac{- ( - 1 ) - 5}{2 \cdot 1} : - 2

二次equationの解:

x = 3, x = - 2