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受欢迎的代数 >

n(t)=15t^2-180t+5000

解答

n (t) = 15 t^{2} - 180 t + 5000

解答

t = \frac{180 + n + n ^{2} + 360 n - 267600}{30}, t = \frac{180 + n - n ^{2} + 360 n - 267600}{30}

求解步骤

n (t) = 15 t^{2} - 180 t + 5000

整理后得

n t = 15 t^{2} - 180 t + 5000

交换两边

15 t^{2} - 180 t + 5000 = n t

将

n t

para o lado esquerdo

15 t^{2} - 180 t + 5000 - n t = 0

改写成标准形式

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

15 t^{2} - (180 + n) t + 5000 = 0

使用求根公式求解

t_{1, 2} = \frac{- ( - 180 - n ) \pm ( - 180 - n ) ^{2} - 4 \cdot 15 \cdot 5000}{2 \cdot 15}

化简

(- 180 - n)^{2} - 4 \cdot 15 \cdot 5000 : n^{2} + 360 n - 267600

t_{1, 2} = \frac{- ( - 180 - n ) \pm n ^{2} + 360 n - 267600}{2 \cdot 15}

将解分隔开

t_{1} = \frac{- ( - 180 - n ) + n ^{2} + 360 n - 267600}{2 \cdot 15}, t_{2} = \frac{- ( - 180 - n ) - n ^{2} + 360 n - 267600}{2 \cdot 15}

t = \frac{- ( - 180 - n ) + n ^{2} + 360 n - 267600}{2 \cdot 15} : \frac{180 + n + n ^{2} + 360 n - 267600}{30}

t = \frac{- ( - 180 - n ) - n ^{2} + 360 n - 267600}{2 \cdot 15} : \frac{180 + n - n ^{2} + 360 n - 267600}{30}

二次方程组的解是:

t = \frac{180 + n + n ^{2} + 360 n - 267600}{30}, t = \frac{180 + n - n ^{2} + 360 n - 267600}{30}

作图

流行的例子

6x^2+(a-12)x-(a^2+a-6)=0

6 x^{2} + (a - 12) x - (a^{2} + a - 6) = 0

4x^2-4a^2x+(a^4-b^4)=0

4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0

solvefor x,p^2+q^2=x^2+y^2

so l v e f or x, p^{2} + q^{2} = x^{2} + y^{2}

solvefor y,x^3-x*y^2+3y^2+2=0

so l v e f or y, x^{3} - x \cdot y^{2} + 3 y^{2} + 2 = 0

solvefor x,f(1+x)(x-5)=y

so l v e f or x, f (1 + x) (x - 5) = y