2r^2+5r-3=0

解答

2 r^{2} + 5 r - 3 = 0

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

十进制

r = 0.5, r = - 3

求解步骤

2 r^{2} + 5 r - 3 = 0

使用求根公式求解

r_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 3) = 7

r_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 2}

将解分隔开

r_{1} = \frac{- 5 + 7}{2 \cdot 2}, r_{2} = \frac{- 5 - 7}{2 \cdot 2}

r = \frac{- 5 + 7}{2 \cdot 2} : \frac{1}{2}

r = \frac{- 5 - 7}{2 \cdot 2} : - 3

二次方程组的解是:

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

s im pl i f y (7 x - 3 y) (2 x - 3 y) - (3 y - 2 x)^{2}

\frac{∣ x - 1 ∣}{3 x - 1} \geq 0

f a c t or 8 x^{- \frac{1}{2}} + 6 x^{\frac{1}{2}} + x^{\frac{3}{2}}

- 1 < \frac{n}{2} + 1

f a c t or 5 x^{3} + 6 x^{2} + 9 x