(x+2)(x²-2x+4)=7

To solve the equation \((x + 2)(x^2 - 2x + 4) = 7\), we can start by expanding the left side:

1. Expand the expression \((x + 2)(x^2 - 2x + 4)\):

\[ (x + 2)(x^2 - 2x + 4) = x(x^2 - 2x + 4) + 2(x^2 - 2x + 4) \] \[ = x^3 - 2x^2 + 4x + 2x^2 - 4x + 8 \] \[ = x^3 + 8 \]

2. Now set the equation equal to 7:

\[ x^3 + 8 = 7 \]

3. Subtract 7 from both sides:

\[ x^3 + 8 - 7 = 0 \] \[ x^3 + 1 = 0 \]

4. This can be factored as:

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

5. Now, set each factor equal to zero:

• For \(x + 1 = 0\): \[ x = -1 \]

• For \(x^2 - x + 1 = 0\), we use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

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

Calculate the discriminant: \[ b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3 \]

Since the discriminant is negative, the quadratic equation has no real roots. The roots are complex and are given as:

\[ x = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2} \]

6. Thus, the solutions to the original equation \((x + 2)(x^2 - 2x + 4) = 7\) are:

• \(x = -1\)
• \(x = \frac{1 + i\sqrt{3}}{2}\)
• \(x = \frac{1 - i\sqrt{3}}{2}\)

So the complete set of solutions is:

\[ \boxed{-1, \frac{1 + i\sqrt{3}}{2}, \frac{1 - i\sqrt{3}}{2}} \]