-t^2+2t+8=0

解答

- t^{2} + 2 t + 8 = 0

t = - 2, t = 4

求解步骤

- t^{2} + 2 t + 8 = 0

使用求根公式求解

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) \cdot 8}{2 ( - 1 )}

2^{2} - 4 (- 1) \cdot 8 = 6

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

将解分隔开

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

t = \frac{- 2 + 6}{2 ( - 1 )} : - 2

t = \frac{- 2 - 6}{2 ( - 1 )} : 4

二次方程组的解是:

t = - 2, t = 4

so l v e f or x, 4 x^{2} - 4 x + 4 y^{2} - k + 20 = 0

so l v e f or y, x^{2} + (y - 3 x)^{2} = 1

- x^{2} + 14 = 10

x^{2} - 10 x - 19 = 0

(3 x - 5) (7 - 5 x) + (5 x + 2) (2 x - 2) - 2 = 0