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We are tasked with solving the equation:

\[
3^{2x+1} + 8(3^x) - 3 = 0
\]

### Step 1: Express the powers of 3 in terms of \( 3^x \)

We know that \( 3^{2x+1} \) can be rewritten as:

\[
3^{2x+1} = 3 \cdot 3^{2x} = 3 \cdot (3^x)^2
\]

So, the original equation becomes:

\[
3 \cdot (3^x)^2 + 8(3^x) - 3 = 0
\]

### Step 2: Substitution

Let \( y = 3^x \). This substitution simplifies the equation to:

\[
3y^2 + 8y - 3 = 0
\]

### Step 3: Solve the quadratic equation

We now solve the quadratic equation \( 3y^2 + 8y - 3 = 0 \) using the quadratic formula. The quadratic formula is given by:

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

For our equation, \( a = 3 \), \( b = 8 \), and \( c = -3 \). Plugging these values into the formula:

\[
y = \frac{-8 \pm \sqrt{8^2 - 4(3)(-3)}}{2(3)}
\]
\[
y = \frac{-8 \pm \sqrt{64 + 36}}{6}
\]
\[
y = \frac{-8 \pm \sqrt{100}}{6}
\]
\[
y = \frac{-8 \pm 10}{6}
\]

### Step 4: Find the roots

This gives us two possible solutions for \( y \):

1. \( y = \frac{-8 + 10}{6} = \frac{2}{6} = \frac{1}{3} \)
2. \( y = \frac{-8 - 10}{6} = \frac{-18}{6} = -3 \)

### Step 5: Solve for \( x \)

Since \( y = 3^x \), we now solve for \( x \).

#### Case 1: \( y = \frac{1}{3} \)

\[
3^x = \frac{1}{3}
\]

This implies:

\[
x = -1
\]

#### Case 2: \( y = -3 \)

There is no real solution for \( 3^x = -3 \) because the exponential function \( 3^x \) is always positive. Therefore, this case has no solution.

### Final Answer:

The only solution is \( x = -1 \).

---

Would you like further clarification on any step?

Here are some related questions:
1. How does the quadratic formula work in other types of equations?
2. Can exponential functions ever have negative values?
3. How can we check if a solution is correct for an exponential equation?
4. What is the process of substitution in solving algebraic equations?
5. How does rewriting terms like \( 3^{2x+1} \) help in solving exponential equations?

**Tip**: Always check if a quadratic equation is factorable before using the quadratic formula—it can save time!

Solve the equation 3^(2x+1) + 8(3^x) - 3 = 0

Learn how to solve the exponential equation 3^(2x+1) + 8(3^x) - 3 = 0 using substitution and the quadratic formula. This problem involves algebraic concepts such as rewriting exponential terms and solving quadratic equations. Suitable for students in Grades 9-12.

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