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To solve \( 5i(5 - 3i) \), we will use the distributive property to expand the expression.

### Solution
1. Distribute \( 5i \) to both terms inside the parentheses:
   \[
   5i \cdot 5 - 5i \cdot 3i
   \]

2. Calculate each term separately:
   - \( 5i \cdot 5 = 25i \)
   - \( 5i \cdot 3i = 15i^2 \)

3. Substitute \( i^2 = -1 \) (since \( i \) is the imaginary unit):
   \[
   15i^2 = 15(-1) = -15
   \]

4. Substitute back and simplify:
   \[
   25i - 15
   \]

5. Rearrange the terms to standard form for complex numbers:
   \[
   -15 + 25i
   \]

### Final Answer
\[
-15 + 25i
\]

Would you like further details or have any questions?

### Related Questions
1. What is the value of \( i^3 \) and \( i^4 \) in complex numbers?
2. How do you multiply two complex numbers like \( (2 + 3i)(4 - i) \)?
3. What does it mean for a complex number to be in "standard form"?
4. How would you divide two complex numbers?
5. How can you find the modulus (magnitude) of a complex number?

### Tip
Remember that \( i^2 = -1 \) is fundamental in working with complex numbers; it allows you to convert powers of \( i \) into real numbers.

Learn how to multiply complex numbers using the example of 5i (5 - 3i). This detailed solution demonstrates the use of the distributive property and imaginary unit properties to arrive at the answer -15 + 25i, suitable for Grades 10-12 students.

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