1. **State the problem:** Solve the quadratic equation $$2x^3 - 2x^1 3 - 8 = 0$$. It appears there might be a formatting issue, so we interpret the equation as $$2x^3 - 2x - 8 = 0$$. 2. **Rewrite the equation:** $$2x^3 - 2x - 8 = 0$$. 3. **Divide the entire equation by 2** to simplify: $$x^3 - x - 4 = 0$$. 4. **Try to find rational roots using the Rational Root Theorem:** Possible roots are factors of 4: $$\pm1, \pm2, \pm4$$. 5. **Test $x=2$:** $$2^3 - 2 - 4 = 8 - 2 - 4 = 2 \neq 0$$. 6. **Test $x=1$:** $$1^3 - 1 - 4 = 1 - 1 - 4 = -4 \neq 0$$. 7. **Test $x=-1$:** $$(-1)^3 - (-1) - 4 = -1 + 1 - 4 = -4 \neq 0$$. 8. **Test $x=-2$:** $$(-2)^3 - (-2) - 4 = -8 + 2 - 4 = -10 \neq 0$$. 9. **Test $x=4$:** $$4^3 - 4 - 4 = 64 - 4 - 4 = 56 \neq 0$$. 10. **Test $x=-4$:** $$(-4)^3 - (-4) - 4 = -64 + 4 - 4 = -64 \neq 0$$. 11. Since no rational roots are found, use the **Cardano's formula** for cubic equations of the form $$x^3 + px + q = 0$$, where here $$p = -1$$ and $$q = -4$$. 12. **Calculate the discriminant:** $$\Delta = \left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3 = \left(-2\right)^2 + \left(-\frac{1}{3}\right)^3 = 4 - \frac{1}{27} = \frac{108}{27} - \frac{1}{27} = \frac{107}{27} > 0$$. 13. Since $$\Delta > 0$$, there is one real root and two complex conjugate roots. 14. **Calculate:** $$u = \sqrt[3]{-\frac{q}{2} + \sqrt{\Delta}} = \sqrt[3]{2 + \sqrt{\frac{107}{27}}}$$ $$v = \sqrt[3]{-\frac{q}{2} - \sqrt{\Delta}} = \sqrt[3]{2 - \sqrt{\frac{107}{27}}}$$ 15. **The real root is:** $$x = u + v$$ 16. **Approximate the root:** $$\sqrt{\frac{107}{27}} \approx \sqrt{3.96296} \approx 1.991$$ $$u = \sqrt[3]{2 + 1.991} = \sqrt[3]{3.991} \approx 1.587$$ $$v = \sqrt[3]{2 - 1.991} = \sqrt[3]{0.009} \approx 0.208$$ 17. **Sum:** $$x \approx 1.587 + 0.208 = 1.795$$ 18. **The two complex roots** can be found using the cube roots of unity but are not requested here. **Final answer:** $$x \approx 1.795$$