Quadratic Equation 2E4Fd8

1. **State the problem:** Solve the quadratic equation $$2x^2 - 4x + 5 = 0$$. 2. **Formula used:** The quadratic formula for solving $$ax^2 + bx + c = 0$$ is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 3. **Identify coefficients:** Here, $$a = 2$$, $$b = -4$$, and $$c = 5$$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 2 \times 5 = 16 - 40 = -24$$. 5. Since $$\Delta < 0$$, the equation has two complex (imaginary) solutions. 6. **Apply the quadratic formula:** $$x = \frac{-(-4) \pm \sqrt{-24}}{2 \times 2} = \frac{4 \pm \sqrt{-24}}{4}$$ 7. Simplify the square root of the negative number: $$\sqrt{-24} = \sqrt{-1 \times 24} = i\sqrt{24} = i\sqrt{4 \times 6} = 2i\sqrt{6}$$ 8. Substitute back: $$x = \frac{4 \pm 2i\sqrt{6}}{4}$$ 9. Simplify the fraction by dividing numerator and denominator by 2: $$x = \frac{\cancel{2} \times 2 \pm \cancel{2} i\sqrt{6}}{\cancel{2} \times 2} = \frac{2 \pm i\sqrt{6}}{2}$$ 10. **Final solutions:** $$x = 1 \pm \frac{i\sqrt{6}}{2}$$ These are the two complex roots of the quadratic equation.