The equation shown in the image is:

$25^{x^2 + 2x + 2} = 125^{x - 2}$

We can solve this by expressing both sides as powers of a common base. Notice that both 25 and 125 are powers of 5:

$25 = 5^2 \quad \text{and} \quad 125 = 5^3$

Now, rewrite the equation using these expressions:

$(5^2)^{x^2 + 2x + 2} = (5^3)^{x - 2}$

Simplify the exponents:

$5^{2(x^2 + 2x + 2)} = 5^{3(x - 2)}$

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

$2(x^2 + 2x + 2) = 3(x - 2)$

Expand both sides:

$2x^2 + 4x + 4 = 3x - 6$

Now, move all terms to one side:

$2x^2 + 4x + 4 - 3x + 6 = 0$

Simplify:

$2x^2 + x + 10 = 0$

Now we can solve this quadratic equation using the quadratic formula:

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

For the equation $2x^2 + x + 10 = 0$, the coefficients are:

$a = 2, \quad b = 1, \quad c = 10$

Let's calculate the discriminant:

$\Delta = b^2 - 4ac = 1^2 - 4(2)(10) = 1 - 80 = -79$

Since the discriminant is negative, the solutions will be complex. Using the quadratic formula:

$x = \frac{-1 \pm \sqrt{-79}}{4}$

$x = \frac{-1 \pm \sqrt{79}i}{4}$

Thus, the solutions are:

$x = \frac{-1 + \sqrt{79}i}{4} \quad \text{or} \quad x = \frac{-1 - \sqrt{79}i}{4}$

These are the two complex solutions to the equation.

Do you have any questions or want more details?

Here are some related questions:

1. What is the quadratic formula, and when is it used?
2. How do you solve a quadratic equation when the discriminant is negative?
3. What are complex numbers and how are they represented?
4. Can this equation have any real solutions?
5. How do exponents with common bases help in solving equations?

Tip: When solving equations, always look for ways to simplify by finding common bases or factoring to make the process easier.