-16t^2+116t+101=0

解答

- 16 t^{2} + 116 t + 101 = 0

t = - \frac{- 29 + 1245}{8}, t = \frac{29 + 1245}{8}

十进制

t = - 0.78556 \dots, t = 8.03556 \dots

求解步骤

- 16 t^{2} + 116 t + 101 = 0

使用求根公式求解

t_{1, 2} = \frac{- 116 \pm 11 6 ^{2} - 4 ( - 16 ) \cdot 101}{2 ( - 16 )}

11 6^{2} - 4 (- 16) \cdot 101 = 41245

t_{1, 2} = \frac{- 116 \pm 4 1245}{2 ( - 16 )}

将解分隔开

t_{1} = \frac{- 116 + 4 1245}{2 ( - 16 )}, t_{2} = \frac{- 116 - 4 1245}{2 ( - 16 )}

t = \frac{- 116 + 4 1245}{2 ( - 16 )} : - \frac{- 29 + 1245}{8}

t = \frac{- 116 - 4 1245}{2 ( - 16 )} : \frac{29 + 1245}{8}

二次方程组的解是:

t = - \frac{- 29 + 1245}{8}, t = \frac{29 + 1245}{8}

(12 y - 2 (\frac{18 - 3 y}{4}) y - 3 y^{2}) = 0

f a c t or 2 x^{2} - 9 x = 0

so l v e f or x, x^{2} y + 2 y^{3} = 3 x + 2 y

4 n^{2} - 8 n + 90 = 9

so l v e f or x, h + g = x^{2}