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.