6sqrt(3)*x^2+7x=sqrt(3)

解答

63 \cdot x^{2} + 7 x = 3

x = \frac{1}{3 3}, x = - \frac{3}{2}

十进制

x = 0.19245 \dots, x = - 0.86602 \dots

求解步骤

63 x^{2} + 7 x = 3

将

3

para o lado esquerdo

63 x^{2} + 7 x - 3 = 0

使用求根公式求解

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 6 3 ( - 3 )}{2 \cdot 6 3}

7^{2} - 4 \cdot 63 (- 3) = 11

x_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 6 3}

将解分隔开

x_{1} = \frac{- 7 + 11}{2 \cdot 6 3}, x_{2} = \frac{- 7 - 11}{2 \cdot 6 3}

x = \frac{- 7 + 11}{2 \cdot 6 3} : \frac{1}{3 3}

x = \frac{- 7 - 11}{2 \cdot 6 3} : - \frac{3}{2}

二次方程组的解是:

x = \frac{1}{3 3}, x = - \frac{3}{2}

so l v e f or n, p = (n^{2}) + 1

(\frac{x + 1}{3}) (\frac{x - 1}{3}) = \frac{1}{3}

so l v e f or x, z = 5 x_{3} x^{2} y^{+} 7 y^{5}

x^{3} - 4 x^{2} - x + 1 = (x - 2)^{3}

so l v e f or x, y^{2} - 1 = (x - 1) (x - 3)