求解 2x^3+4x^2-8x+13=0

$$2 x ^ { 3 } + 4 x ^ { 2 } - 8 x + 13 = 0$$

求解 x 的值 (复数求解)

$$x=-\frac{4^{\frac{2}{3}}\left(1+\sqrt{3}i\right)\left(-\sqrt{3}\sqrt[3]{3\sqrt{23577}-527}i-2\sqrt{3}\sqrt[3]{4}i-\sqrt[3]{3\sqrt{23577}-527}+2\sqrt[3]{-3\sqrt{23577}-527}+2\sqrt[3]{4}\right)}{48}\approx -0.471204465-1.811018443i$$

$$x=\frac{4^{\frac{2}{3}}\left(\sqrt[3]{3\sqrt{23577}-527}+\sqrt[3]{-3\sqrt{23577}-527}-2\sqrt[3]{4}\right)}{12}\approx 0.803990211+2.547252433i$$

$$x=-\frac{4^{\frac{2}{3}}\left(-\sqrt{3}i+1\right)\left(-\sqrt[3]{3\sqrt{23577}-527}+2\sqrt[3]{-3\sqrt{23577}-527}+2\sqrt[3]{4}+\sqrt{3}\sqrt[3]{3\sqrt{23577}-527}i+2\sqrt{3}\sqrt[3]{4}i\right)}{48}\approx -2.332785746-0.736233989i$$

求解 x 的值

$$x=-\frac{4^{\frac{2}{3}}\sqrt[3]{3\sqrt{23577}+527}}{12}-\frac{4^{\frac{2}{3}}\sqrt[3]{527-3\sqrt{23577}}}{12}-\frac{2}{3}\approx -3.607980422$$