(х2+2х+2)(х2+2х-4)=-5

To solve the equation \((x^2 + 2x + 2)(x^2 + 2x - 4) = -5\), we first set:

Let \(y = x^2 + 2x\). Then, the equation can be rewritten as:

\[ (y + 2)(y - 4) = -5 \]

Now, expanding the left-hand side gives us:

\[ y^2 - 4y + 2y - 8 = y^2 - 2y - 8 \]

Setting the equation equal to \(-5\):

\[ y^2 - 2y - 8 = -5 \]

Now, add 5 to both sides:

\[ y^2 - 2y - 3 = 0 \]

Next, we can solve this quadratic equation using the quadratic formula:

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

where \(a = 1\), \(b = -2\), and \(c = -3\):

\[ y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} \] \[ y = \frac{2 \pm \sqrt{4 + 12}}{2} \] \[ y = \frac{2 \pm \sqrt{16}}{2} \] \[ y = \frac{2 \pm 4}{2} \]

This results in two solutions for \(y\):

1. \(y = \frac{6}{2} = 3\)
2. \(y = \frac{-2}{2} = -1\)

Now, we will replace \(y\) back with \(x^2 + 2x\) and solve for \(x\):

1. For \(y = 3\):

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

Factoring gives:

\[ (x + 3)(x - 1) = 0 \]

So, \(x = -3\) or \(x = 1\).

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

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

This factors to:

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

So, \(x = -1\) (with a multiplicity of 2).

Final Solutions

The solutions to the original equation \((x^2 + 2x + 2)(x^2 + 2x - 4) = -5\) are:

\[ x = -3, \quad x = 1, \quad x = -1 \text{ (with multiplicity 2)} \]