r^2-13r+42=0

解答

r^{2} - 13 r + 42 = 0

r = 7, r = 6

求解步骤

r^{2} - 13 r + 42 = 0

使用求根公式求解

r_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 42 = 1

r_{1, 2} = \frac{- ( - 13 ) \pm 1}{2 \cdot 1}

将解分隔开

r_{1} = \frac{- ( - 13 ) + 1}{2 \cdot 1}, r_{2} = \frac{- ( - 13 ) - 1}{2 \cdot 1}

r = \frac{- ( - 13 ) + 1}{2 \cdot 1} : 7

r = \frac{- ( - 13 ) - 1}{2 \cdot 1} : 6

二次方程组的解是:

r = 7, r = 6

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