21T^2-320T+448=0

解

21 T^{2} - 320 T + 448 = 0

T = \frac{8 ( 20 + 253 )}{21}, T = \frac{8 ( 20 - 253 )}{21}

十進法表記

T = 13.67846 \dots, T = 1.55962 \dots

解答ステップ

21 T^{2} - 320 T + 448 = 0

解くとthe二次式

T_{1, 2} = \frac{- ( - 320 ) \pm ( - 320 ) ^{2} - 4 \cdot 21 \cdot 448}{2 \cdot 21}

(- 320)^{2} - 4 \cdot 21 \cdot 448 = 16253

T_{1, 2} = \frac{- ( - 320 ) \pm 16 253}{2 \cdot 21}

解を分離する

T_{1} = \frac{- ( - 320 ) + 16 253}{2 \cdot 21}, T_{2} = \frac{- ( - 320 ) - 16 253}{2 \cdot 21}

T = \frac{- ( - 320 ) + 16 253}{2 \cdot 21} : \frac{8 ( 20 + 253 )}{21}

T = \frac{- ( - 320 ) - 16 253}{2 \cdot 21} : \frac{8 ( 20 - 253 )}{21}

二次equationの解:

T = \frac{8 ( 20 + 253 )}{21}, T = \frac{8 ( 20 - 253 )}{21}