(х+3)^4-2(х+3)^2-15=0

To solve the equation \((x + 3)^4 - 2(x + 3)^2 - 15 = 0\), we can use a substitution method. Let's set:

\[ y = (x + 3)^2 \]

Now, the equation becomes:

\[ y^2 - 2y - 15 = 0 \]

Next, we can factor the quadratic equation. We are looking for two numbers that multiply to \(-15\) and add up to \(-2\). The numbers \(-5\) and \(3\) satisfy this condition. Therefore, we can factor the equation as follows:

\[ (y - 5)(y + 3) = 0 \]

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

1. \(y - 5 = 0 \Rightarrow y = 5\)
2. \(y + 3 = 0 \Rightarrow y = -3\)

Now, we substitute back \(y = (x + 3)^2\):

1. For \(y = 5\): \[ (x + 3)^2 = 5 \] Taking the square root of both sides, we get: \[ x + 3 = \pm \sqrt{5} \] Therefore: \[ x = -3 + \sqrt{5} \quad \text{and} \quad x = -3 - \sqrt{5} \]

2. For \(y = -3\): \[ (x + 3)^2 = -3 \] Since the square of a real number cannot be negative, this case has no real solutions.

Thus, the real solutions to the original equation are:

\[ x = -3 + \sqrt{5} \quad \text{and} \quad x = -3 - \sqrt{5} \]