https://r2.erweima.ai/mathbot/image/1726615512397oz3h6bgm.jpg

The equation shown in the image is:

\[
3^{x^2 - 3x - 1} = 9^{x + 2}
\]

We will start by simplifying the equation. Notice that \(9\) can be written as \(3^2\), so the equation becomes:

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

This simplifies further to:

\[
3^{x^2 - 3x - 1} = 3^{2(x + 2)} = 3^{2x + 4}
\]

Since both sides of the equation have the same base (3), we can equate the exponents:

\[
x^2 - 3x - 1 = 2x + 4
\]

Now, let's solve this quadratic equation by moving all terms to one side:

\[
x^2 - 3x - 1 - 2x - 4 = 0
\]

Simplify the equation:

\[
x^2 - 5x - 5 = 0
\]

We now solve this quadratic equation using the quadratic formula:

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

For the equation \(x^2 - 5x - 5 = 0\), \(a = 1\), \(b = -5\), and \(c = -5\). Substituting these values into the formula:

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-5)}}{2(1)}
\]
\[
x = \frac{5 \pm \sqrt{25 + 20}}{2}
\]
\[
x = \frac{5 \pm \sqrt{45}}{2}
\]
\[
x = \frac{5 \pm 3\sqrt{5}}{2}
\]

Thus, the two possible solutions for \(x\) are:

\[
x = \frac{5 + 3\sqrt{5}}{2} \quad \text{or} \quad x = \frac{5 - 3\sqrt{5}}{2}
\]

These are the solutions to the given equation.

Would you like more details on solving quadratic equations or have any other questions?

### Related Questions:
1. How do you simplify expressions with exponents having the same base?
2. What is the process for expanding binomials in quadratic equations?
3. Can the quadratic formula always be used for solving second-degree equations?
4. How does factoring compare to the quadratic formula for solving quadratics?
5. Why do quadratic equations sometimes have complex or irrational solutions?

### Tip:
When equating exponents, always ensure both sides have the same base to directly compare the exponents easily.

Discover how to solve the equation 3^(x^2 - 3x - 1) = 9^(x + 2) by simplifying exponents and using the quadratic formula. This step-by-step guide explains how to equate exponents and solve quadratic equations, suitable for high school students in Grades 10-12.

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