We are tasked with solving the equation:

\[
e^{2x} + 5e^x - 36 = 0
\]

### Step-by-Step Solution:

1. **Substitute**: Let \( u = e^x \), which simplifies the equation to:
   \[
   u^2 + 5u - 36 = 0
   \]
   This is now a quadratic equation in terms of \( u \).

2. **Solve the quadratic equation**: We can use the quadratic formula:
   \[
   u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = 1 \), \( b = 5 \), and \( c = -36 \).

   Plugging in the values:
   \[
   u = \frac{-5 \pm \sqrt{5^2 - 4(1)(-36)}}{2(1)}
   \]
   \[
   u = \frac{-5 \pm \sqrt{25 + 144}}{2}
   \]
   \[
   u = \frac{-5 \pm \sqrt{169}}{2}
   \]
   \[
   u = \frac{-5 \pm 13}{2}
   \]

   So, we have two solutions for \( u \):
   \[
   u_1 = \frac{-5 + 13}{2} = 4
   \]
   \[
   u_2 = \frac{-5 - 13}{2} = -9
   \]

3. **Back-substitute \( u = e^x \)**:
   - For \( u_1 = 4 \): 
     \[
     e^x = 4 \implies x = \ln(4)
     \]
   - For \( u_2 = -9 \): 
     \( e^x \) cannot be negative (since the exponential function is always positive), so this solution is not valid.

### Final Solution:
The only valid solution is:
\[
x = \ln(4)
\]

Would you like more details or have any questions?

### Related Questions:
1. How do you solve quadratic equations using factoring?
2. What is the quadratic formula, and when do you use it?
3. How do you solve exponential equations using logarithms?
4. Can exponential functions have negative outputs?
5. What are some real-world applications of exponential equations?

**Tip**: When solving exponential equations, substituting \( u = e^x \) can simplify the problem into a manageable quadratic form.

Discover how to solve the exponential equation e^2x + 5e^x - 36 = 0 by using a quadratic substitution technique. This step-by-step solution covers key algebraic concepts, including the quadratic formula and logarithms. Suitable for high school students in Grades 10-12.

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