Let's solve the given equation step by step.

### Given equation:
\[
5x^2 - 13x + 15 = -5x
\]

### Step 1: Move all terms to one side to set the equation equal to 0.
\[
5x^2 - 13x + 5x + 15 = 0
\]
Simplify:
\[
5x^2 - 8x + 15 = 0
\]

### Step 2: Apply the quadratic formula.
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For the equation \( 5x^2 - 8x + 15 = 0 \), the coefficients are:
- \( a = 5 \)
- \( b = -8 \)
- \( c = 15 \)

Substitute these values into the quadratic formula:
\[
x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(15)}}{2(5)}
\]
Simplify:
\[
x = \frac{8 \pm \sqrt{64 - 300}}{10}
\]
\[
x = \frac{8 \pm \sqrt{-236}}{10}
\]

Since the discriminant is negative (\( \sqrt{-236} \)), the solutions are complex numbers.

### Step 3: Express the complex solution.
\[
x = \frac{8 \pm \sqrt{236}i}{10}
\]
Simplify further:
\[
x = \frac{8}{10} \pm \frac{\sqrt{236}}{10}i
\]
\[
x = 0.8 \pm 0.485i
\]

### Final Answer:
The two complex solutions are:
\[
x = 0.8 + 0.485i \quad \text{and} \quad x = 0.8 - 0.485i
\]

Let me know if you'd like more details or have any questions!

Here are five related questions for further practice:
1. Solve \( 4x^2 - 12x + 9 = 0 \).
2. Factorize \( 3x^2 - 5x - 2 = 0 \).
3. What happens if the discriminant of a quadratic is zero?
4. Solve \( 2x^2 + 3x + 5 = 0 \) using the quadratic formula.
5. What are the real-life applications of complex numbers in physics?

**Tip:** If the discriminant is negative, the quadratic equation will have complex solutions!

Learn how to solve the quadratic equation 5x^2 − 13x + 15 = −5x using the quadratic formula. This problem involves algebraic concepts such as quadratic equations and complex numbers, leading to solutions in the form of complex numbers. Suitable for students in Grades 9-12.

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