4n^2-20n+16=0

解

4 n^{2} - 20 n + 16 = 0

n = 4, n = 1

解答ステップ

4 n^{2} - 20 n + 16 = 0

解くとthe二次式

n_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

(- 20)^{2} - 4 \cdot 4 \cdot 16 = 12

n_{1, 2} = \frac{- ( - 20 ) \pm 12}{2 \cdot 4}

解を分離する

n_{1} = \frac{- ( - 20 ) + 12}{2 \cdot 4}, n_{2} = \frac{- ( - 20 ) - 12}{2 \cdot 4}

n = \frac{- ( - 20 ) + 12}{2 \cdot 4} : 4

n = \frac{- ( - 20 ) - 12}{2 \cdot 4} : 1

二次equationの解:

n = 4, n = 1