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受欢迎的代数 >

N(d)=-2d^2+16d-14

解答

N (d) = - 2 d^{2} + 16 d - 14

解答

d = - \frac{- 16 + N + N ^{2} - 32 N + 144}{4}, d = - \frac{- 16 + N - N ^{2} - 32 N + 144}{4}

求解步骤

N (d) = - 2 d^{2} + 16 d - 14

整理后得

N d = - 2 d^{2} + 16 d - 14

交换两边

- 2 d^{2} + 16 d - 14 = N d

将

N d

para o lado esquerdo

- 2 d^{2} + 16 d - 14 - N d = 0

改写成标准形式

a x^{2} + b x + c = 0

- 2 d^{2} + (16 - N) d - 14 = 0

使用求根公式求解

d_{1, 2} = \frac{- ( 16 - N ) \pm ( 16 - N ) ^{2} - 4 ( - 2 ) ( - 14 )}{2 ( - 2 )}

化简

(16 - N)^{2} - 4 (- 2) (- 14) : N^{2} - 32 N + 144

d_{1, 2} = \frac{- ( 16 - N ) \pm N ^{2} - 32 N + 144}{2 ( - 2 )}

将解分隔开

d_{1} = \frac{- ( 16 - N ) + N ^{2} - 32 N + 144}{2 ( - 2 )}, d_{2} = \frac{- ( 16 - N ) - N ^{2} - 32 N + 144}{2 ( - 2 )}

d = \frac{- ( 16 - N ) + N ^{2} - 32 N + 144}{2 ( - 2 )} : - \frac{- 16 + N + N ^{2} - 32 N + 144}{4}

d = \frac{- ( 16 - N ) - N ^{2} - 32 N + 144}{2 ( - 2 )} : - \frac{- 16 + N - N ^{2} - 32 N + 144}{4}

二次方程组的解是:

d = - \frac{- 16 + N + N ^{2} - 32 N + 144}{4}, d = - \frac{- 16 + N - N ^{2} - 32 N + 144}{4}

作图

流行的例子

0.8x^2-8.16x+4.8=0

0.8 x^{2} - 8.16 x + 4.8 = 0

8 x^{2} - 2 x - 21 = 0

solvefor a,a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)

so l v e f or a, a^{5} + b^{5} = (a + b) (a^{4} - a^{3} b + a^{2} b^{2} - a b^{3} + b^{4})

solvefor q, 1/4 q^2+r=p

so l v e f or q, \frac{1}{4} q^{2} + r = p

a^{2} + 1 3^{2} = 1 2^{2}