1. **State the problem:** Solve the quadratic equation $2x^2 - 4x + 5 = 0$. 2. **Recall the quadratic formula:** For an equation $ax^2 + bx + c = 0$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-4$, and $c=5$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 2 \times 5 = 16 - 40 = -24$$ Since $\Delta < 0$, the equation has no real roots but two complex roots. 4. **Apply the quadratic formula:** $$x = \frac{-(-4) \pm \sqrt{-24}}{2 \times 2} = \frac{4 \pm \sqrt{-24}}{4}$$ 5. **Simplify the square root of the negative number:** $$\sqrt{-24} = \sqrt{-1 \times 24} = i \sqrt{24} = i \sqrt{4 \times 6} = 2i \sqrt{6}$$ 6. **Substitute back:** $$x = \frac{4 \pm 2i \sqrt{6}}{4}$$ 7. **Simplify the fraction by canceling common factors:** $$x = \frac{\cancel{2} \times 2 \pm \cancel{2} i \sqrt{6}}{\cancel{2} \times 2} = \frac{2 \pm i \sqrt{6}}{2}$$ 8. **Final answer:** $$x = 1 \pm \frac{i \sqrt{6}}{2}$$ This means the solutions are complex conjugates: $x = 1 + \frac{i \sqrt{6}}{2}$ and $x = 1 - \frac{i \sqrt{6}}{2}$.