x(x+1)(x+2)(x+3)=24

To solve the equation \( x(x+1)(x+2)(x+3) = 24 \), we can start by letting \( y = x(x+1)(x+2)(x+3) \).

This equation is a polynomial expression of degree 4, meaning it could have up to 4 real roots. However, to make it easier to handle, we can start by simplifying the left side.

Notice that:

\[ x(x+1)(x+2)(x+3) = (x^2 + 3x)(x^2 + 3x + 2) \]

Let \( z = x^2 + 3x \). Then the equation becomes:

\[ z(z + 2) = 24 \]

Expanding this gives:

\[ z^2 + 2z - 24 = 0 \]

Now, we can use the quadratic formula to solve for \( z \):

\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2} = \frac{-2 \pm 10}{2} \]

Calculating the two possible values for \( z \):

1. \( z = \frac{8}{2} = 4 \)
2. \( z = \frac{-12}{2} = -6 \)

Now, we revert back to \( x \) using \( z = x^2 + 3x \):

1. For \( z = 4 \): \[ x^2 + 3x - 4 = 0 \] Using the quadratic formula: \[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \] This gives us:

   • \( x = \frac{2}{2} = 1 \)
   • \( x = \frac{-8}{2} = -4 \)

2. For \( z = -6 \): \[ x^2 + 3x + 6 = 0 \] Using the quadratic formula again: \[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(6)}}{2(1)} = \frac{-3 \pm \sqrt{9 - 24}}{2} = \frac{-3 \pm \sqrt{-15}}{2} \] Since this gives a negative discriminant, we have no real solutions from this equation.

Thus, the real solutions to the original equation \( x(x+1)(x+2)(x+3) = 24 \) are:

\[ \boxed{1} \quad \text{and} \quad \boxed{-4} \]