6x^2+(2n+3)x+n=0

解

6 x^{2} + (2 n + 3) x + n = 0

x = - \frac{1}{2}, x = - \frac{n}{3}

解答ステップ

6 x^{2} + (2 n + 3) x + n = 0

解くとthe二次式

x_{1, 2} = \frac{- ( 2 n + 3 ) \pm ( 2 n + 3 ) ^{2} - 4 \cdot 6 n}{2 \cdot 6}

簡素化

(2 n + 3)^{2} - 4 \cdot 6 n : 2 n - 3

x_{1, 2} = \frac{- ( 2 n + 3 ) \pm ( 2 n - 3 )}{2 \cdot 6}

解を分離する

x_{1} = \frac{- ( 2 n + 3 ) + 2 n - 3}{2 \cdot 6}, x_{2} = \frac{- ( 2 n + 3 ) - ( 2 n - 3 )}{2 \cdot 6}

x = \frac{- ( 2 n + 3 ) + 2 n - 3}{2 \cdot 6} : - \frac{1}{2}

x = \frac{- ( 2 n + 3 ) - ( 2 n - 3 )}{2 \cdot 6} : - \frac{n}{3}

二次equationの解:

x = - \frac{1}{2}, x = - \frac{n}{3}