x^2+0.13x+0.0032=0

解答

x^{2} + 0.13 x + 0.0032 = 0

x = \frac{- 0.13 + 0.0041}{2}, x = \frac{- 0.13 - 0.0041}{2}

十进制

x = - 0.03298 \dots, x = - 0.09701 \dots

求解步骤

x^{2} + 0.13 x + 0.0032 = 0

使用求根公式求解

x_{1, 2} = \frac{- 0.13 \pm 0.1 3 ^{2} - 4 \cdot 1 \cdot 0.0032}{2 \cdot 1}

0.1 3^{2} - 4 \cdot 1 \cdot 0.0032 = 0.0041

x_{1, 2} = \frac{- 0.13 \pm 0.0041}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- 0.13 + 0.0041}{2 \cdot 1}, x_{2} = \frac{- 0.13 - 0.0041}{2 \cdot 1}

x = \frac{- 0.13 + 0.0041}{2 \cdot 1} : \frac{- 0.13 + 0.0041}{2}

x = \frac{- 0.13 - 0.0041}{2 \cdot 1} : \frac{- 0.13 - 0.0041}{2}

二次方程组的解是:

x = \frac{- 0.13 + 0.0041}{2}, x = \frac{- 0.13 - 0.0041}{2}

so l v e f or y, 2 x^{3} + x^{2} y + y^{2} = 4

0 = 3 x^{2} - 12 x + 12

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

6 x^{2} - 54 = 5 x^{2} + 171

co m pl e t es q u a re x^{2} - 6 x + 9