(х2+2х)2-14(х2+2х)-15=0

To solve the equation \((x^2 + 2x)^2 - 14(x^2 + 2x) - 15 = 0\), we can start by making a substitution. Let:

\[ y = x^2 + 2x \]

Now, we can rewrite the equation as:

\[ y^2 - 14y - 15 = 0 \]

Next, we will solve for \(y\) using the quadratic formula, which is given by:

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

Here, \(a = 1\), \(b = -14\), and \(c = -15\).

Let's calculate the discriminant (\(D\)):

\[ D = b^2 - 4ac = (-14)^2 - 4(1)(-15) = 196 + 60 = 256 \]

Since the discriminant is positive, there will be two real solutions for \(y\).

Now we will apply the quadratic formula:

\[ y = \frac{-(-14) \pm \sqrt{256}}{2(1)} \]

\[ y = \frac{14 \pm 16}{2} \]

Now calculating the two possible values for \(y\):

1. \( y_1 = \frac{14 + 16}{2} = \frac{30}{2} = 15 \)
2. \( y_2 = \frac{14 - 16}{2} = \frac{-2}{2} = -1 \)

Now we have:

1. \( y = 15 \)
2. \( y = -1 \)

Now we substitute back \(y = x^2 + 2x\) for both values.

1. For \(y = 15\):

   \[ x^2 + 2x - 15 = 0 \]

   Factoring this gives:

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

   Thus, \(x = -5\) or \(x = 3\).

2. For \(y = -1\):

   \[ x^2 + 2x + 1 = 0 \]

   This can be factored as:

   \[ (x + 1)(x + 1) = 0 \]

   Thus, \(x = -1\) (a repeated root).

The complete solution set for the original equation is:

\[ x = -5, \quad x = 3, \quad x = -1 \]

These are the solutions for the equation \((x^2 + 2x)^2 - 14(x^2 + 2x) - 15 = 0\).