The given equation is:

\[
25^{x^2} = 5^{3x + 1}
\]

We can express \( 25 \) as \( 5^2 \), which allows us to rewrite the equation:

\[
(5^2)^{x^2} = 5^{3x + 1}
\]

Using the property of exponents \( (a^m)^n = a^{m \cdot n} \), the left-hand side becomes:

\[
5^{2x^2} = 5^{3x + 1}
\]

Since the bases on both sides are the same, we can set the exponents equal to each other:

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

Now, we have a quadratic equation:

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

We can solve this quadratic equation using the quadratic formula:

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

Here, \( a = 2 \), \( b = -3 \), and \( c = -1 \). Plugging these values into the formula:

\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-1)}}{2(2)}
\]

\[
x = \frac{3 \pm \sqrt{9 + 8}}{4}
\]

\[
x = \frac{3 \pm \sqrt{17}}{4}
\]

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

\[
x = \frac{3 + \sqrt{17}}{4} \quad \text{or} \quad x = \frac{3 - \sqrt{17}}{4}
\]

These are the two solutions to the equation.

Would you like more details or have any questions?  
Here are 5 related questions for practice:

1. Solve \( 3^{2x} = 9^{x + 1} \).
2. Solve \( 4^{x+2} = 16^{x} \).
3. Solve \( 2^{x^2} = 16^{x} \).
4. Solve \( 5^{2x} = 125^{x+1} \).
5. Solve the quadratic equation \( 2x^2 + 5x + 3 = 0 \).

**Tip:** When solving exponential equations, try expressing both sides with the same base to simplify the exponents.

Learn how to solve the exponential equation 25^(x^2) = 5^(3x + 1) by applying exponentiation rules and the quadratic formula. This problem involves key algebraic concepts suitable for students in Grades 9-12, with a detailed step-by-step solution.

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