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The equation given is \( 5x^2 - 2x + 3 = 0 \). We can solve this using the quadratic formula:

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

For the equation \( 5x^2 - 2x + 3 = 0 \), the coefficients are:
- \( a = 5 \)
- \( b = -2 \)
- \( c = 3 \)

### Step 1: Calculate the discriminant
The discriminant \( \Delta \) is:

\[
\Delta = b^2 - 4ac = (-2)^2 - 4(5)(3) = 4 - 60 = -56
\]

Since the discriminant is negative, the solutions will be complex.

### Step 2: Apply the quadratic formula
Using the quadratic formula with the discriminant \( \Delta = -56 \):

\[
x = \frac{-(-2) \pm \sqrt{-56}}{2(5)} = \frac{2 \pm \sqrt{-56}}{10}
\]

We can express \( \sqrt{-56} \) as \( \sqrt{56} \cdot i \), where \( i \) is the imaginary unit. Simplifying \( \sqrt{56} \):

\[
\sqrt{56} = \sqrt{4 \cdot 14} = 2\sqrt{14}
\]

So the equation becomes:

\[
x = \frac{2 \pm 2i\sqrt{14}}{10} = \frac{2}{10} \pm \frac{2i\sqrt{14}}{10}
\]

Simplifying:

\[
x = \frac{1}{5} \pm \frac{i\sqrt{14}}{5}
\]

### Final solution:
The solutions are:

\[
x = \frac{1 + i\sqrt{14}}{5} \quad \text{and} \quad x = \frac{1 - i\sqrt{14}}{5}
\]

This corresponds to the third option in the list.

Do you want more details on the steps, or have any other questions?

Here are 5 related questions to consider:
1. How do we interpret the discriminant in terms of real and complex solutions?
2. What happens when the discriminant is zero or positive?
3. Can you explain how to simplify the square root of negative numbers in quadratic equations?
4. What are other methods to solve quadratic equations besides the quadratic formula?
5. How does the value of the coefficient \( a \) affect the shape of the parabola represented by a quadratic equation?

**Tip:** Always check the discriminant first to determine the nature of the roots (real or complex) before solving a quadratic equation.

Solve using the quadratic formula: 5x^2 - 2x + 3 = 0

Learn how to solve the quadratic equation 5x^2 - 2x + 3 = 0 using the quadratic formula. This problem involves algebraic techniques and results in complex solutions, ideal for students in Grades 9-12.

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