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受欢迎的代数 >

-16t^2+70t+1=41

解答

- 16 t^{2} + 70 t + 1 = 41

解答

t = \frac{35 - 3 65}{16}, t = \frac{35 + 3 65}{16}

+1

十进制

t = 0.67582 \dots, t = 3.69917 \dots

求解步骤

- 16 t^{2} + 70 t + 1 = 41

将

41

para o lado esquerdo

- 16 t^{2} + 70 t - 40 = 0

使用求根公式求解

t_{1, 2} = \frac{- 70 \pm 7 0 ^{2} - 4 ( - 16 ) ( - 40 )}{2 ( - 16 )}

7 0^{2} - 4 (- 16) (- 40) = 665

t_{1, 2} = \frac{- 70 \pm 6 65}{2 ( - 16 )}

将解分隔开

t_{1} = \frac{- 70 + 6 65}{2 ( - 16 )}, t_{2} = \frac{- 70 - 6 65}{2 ( - 16 )}

t = \frac{- 70 + 6 65}{2 ( - 16 )} : \frac{35 - 3 65}{16}

t = \frac{- 70 - 6 65}{2 ( - 16 )} : \frac{35 + 3 65}{16}

二次方程组的解是:

t = \frac{35 - 3 65}{16}, t = \frac{35 + 3 65}{16}

作图

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