To solve the quadratic equation:

$2x^2 - 4x + 3 = 0$

we can apply the quadratic formula, which is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, the coefficients are:

• $a = 2$
• $b = -4$
• $c = 3$

Step 1: Calculate the discriminant

The discriminant is the part under the square root in the quadratic formula, calculated as:

$\Delta = b^2 - 4ac$

Substituting the values of $a$, $b$, and $c$:

$\Delta = (-4)^2 - 4(2)(3) = 16 - 24 = -8$

Step 2: Analyze the discriminant

Since the discriminant is negative ($\Delta = -8$), there are no real solutions, but we have two complex solutions.

Step 3: Calculate the complex solutions

Using the quadratic formula, we now compute the roots:

$x = \frac{-(-4) \pm \sqrt{-8}}{2(2)}$

Simplifying:

$x = \frac{4 \pm \sqrt{-8}}{4}$

Since $\sqrt{-8} = \sqrt{8} \cdot i = 2\sqrt{2}i$, we get:

$x = \frac{4 \pm 2\sqrt{2}i}{4}$

Simplifying further:

$x = 1 \pm \frac{\sqrt{2}i}{2}$

Thus, the two complex solutions are:

$x_1 = 1 + \frac{\sqrt{2}i}{2}, \quad x_2 = 1 - \frac{\sqrt{2}i}{2}$

These are the solutions to the quadratic equation.

Would you like more details on any of the steps, or do you have any further questions?

Here are some related questions you might consider:

1. What are the conditions for real vs complex solutions in quadratic equations?
2. How do you solve a quadratic equation with real roots?
3. What are the applications of complex roots in real-world problems?
4. Can a quadratic equation have only one solution? Under what conditions?
5. How do you find the vertex of a parabola given by a quadratic equation?

Tip: The sign of the discriminant ($b^2 - 4ac$) is a key factor in determining whether the solutions are real, repeated, or complex.