To solve the equation (x² + x)² + 5(x² + x) + 6 = 0, we can use a substitution. Let's substitute a variable, such as y, for (x² + x):

Let y = (x² + x)

Now, we can rewrite the equation in terms of y:

y² + 5y + 6 = 0

This is a quadratic equation in terms of y. To solve it, we can factor it:

(y + 2)(y + 3) = 0

Setting each factor equal to zero, we have:

y + 2 = 0 or y + 3 = 0

Solving for y in each equation:

y = -2 or y = -3

Now, substitute back the value of y:

For y = -2:
x² + x = -2

Rearranging the equation:
x² + x + 2 = 0

This quadratic equation does not factor easily, so we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

For this equation, a = 1, b = 1, and c = 2. Plugging in these values:

x = (-1 ± √(1² - 4(1)(2))) / (2(1))
x = (-1 ± √(1 - 8)) / 2
x = (-1 ± √(-7)) / 2

Since the square root of a negative number is not a real number, there are no real solutions for this case.

For y = -3:
x² + x = -3

Rearranging the equation:
x² + x + 3 = 0

Again, this quadratic equation does not factor easily, so we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

For this equation, a = 1, b = 1, and c = 3. Plugging in these values:

x = (-1 ± √(1² - 4(1)(3))) / (2(1))
x = (-1 ± √(1 - 12)) / 2
x = (-1 ± √(-11)) / 2

Since the square root of a negative number is not a real number, there are no real solutions for this case either.

Therefore, the equation (x² + x)² + 5(x² + x) + 6 = 0 has no real solutions.