q^2-22q+120=0

解

q^{2} - 22 q + 120 = 0

q = 12, q = 10

解答ステップ

q^{2} - 22 q + 120 = 0

解くとthe二次式

q_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot 120}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot 120 = 2

q_{1, 2} = \frac{- ( - 22 ) \pm 2}{2 \cdot 1}

解を分離する

q_{1} = \frac{- ( - 22 ) + 2}{2 \cdot 1}, q_{2} = \frac{- ( - 22 ) - 2}{2 \cdot 1}

q = \frac{- ( - 22 ) + 2}{2 \cdot 1} : 12

q = \frac{- ( - 22 ) - 2}{2 \cdot 1} : 10

二次equationの解:

q = 12, q = 10