16t^2-128t+192=0

解答

16 t^{2} - 128 t + 192 = 0

t = 6, t = 2

求解步骤

16 t^{2} - 128 t + 192 = 0

使用求根公式求解

t_{1, 2} = \frac{- ( - 128 ) \pm ( - 128 ) ^{2} - 4 \cdot 16 \cdot 192}{2 \cdot 16}

(- 128)^{2} - 4 \cdot 16 \cdot 192 = 64

t_{1, 2} = \frac{- ( - 128 ) \pm 64}{2 \cdot 16}

将解分隔开

t_{1} = \frac{- ( - 128 ) + 64}{2 \cdot 16}, t_{2} = \frac{- ( - 128 ) - 64}{2 \cdot 16}

t = \frac{- ( - 128 ) + 64}{2 \cdot 16} : 6

t = \frac{- ( - 128 ) - 64}{2 \cdot 16} : 2

二次方程组的解是:

t = 6, t = 2

12 y^{2} + 55 y + 63 = 0

co m pl e t es q u a re 6 a^{2} - 4 a - 22 = 0

(2 x + 7)^{2} = 49

- x^{2} + 5 - 4 x = 0

so l v e f or x, f (k - 1) = 5 x^{2} + 4 x - 5