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受欢迎的代数 >

16t^2-128t+240=0

解答

16 t^{2} - 128 t + 240 = 0

解答

t = 5, t = 3

求解步骤

16 t^{2} - 128 t + 240 = 0

使用求根公式求解

t_{1, 2} = \frac{- ( - 128 ) \pm ( - 128 ) ^{2} - 4 \cdot 16 \cdot 240}{2 \cdot 16}

(- 128)^{2} - 4 \cdot 16 \cdot 240 = 32

t_{1, 2} = \frac{- ( - 128 ) \pm 32}{2 \cdot 16}

将解分隔开

t_{1} = \frac{- ( - 128 ) + 32}{2 \cdot 16}, t_{2} = \frac{- ( - 128 ) - 32}{2 \cdot 16}

t = \frac{- ( - 128 ) + 32}{2 \cdot 16} : 5

t = \frac{- ( - 128 ) - 32}{2 \cdot 16} : 3

二次方程组的解是:

t = 5, t = 3

作图

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