Identify the given matrix X and the identity matrix I.

The given matrix is:

$$X = \begin{pmatrix} -2 & 1 \\ 2 & 4 \end{pmatrix}$$

For a $2 \times 2$ matrix, the identity matrix $I$ is:

$$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

Multiply the identity matrix by the scalar $\lambda$:

$$\lambda I = \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}$$

Subtract $\lambda I$ from $X$:

$$X - \lambda I = \begin{pmatrix} -2 & 1 \\ 2 & 4 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} = \begin{pmatrix} -2 - \lambda & 1 - 0 \\ 2 - 0 & 4 - \lambda \end{pmatrix} = \begin{pmatrix} -2 - \lambda & 1 \\ 2 & 4 - \lambda \end{pmatrix}$$

The determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad - bc$.
So, for $X - \lambda I$:

$$|X - \lambda I| = (-2 - \lambda)(4 - \lambda) - (1)(2)$$

We are given that $|X - \lambda I| = 0$:

$$(-2 - \lambda)(4 - \lambda) - 2 = 0$$

Expand the product:

$$-8 + 2\lambda - 4\lambda + \lambda^2 - 2 = 0$$

Combine like terms:

$$\lambda^2 - 2\lambda - 10 = 0$$

The quadratic formula for $a\lambda^2 + b\lambda + c = 0$ is $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a=1$, $b=-2$, $c=-10$.

$$\lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-10)}}{2(1)}$$ $$\lambda = \frac{2 \pm \sqrt{4 + 40}}{2}$$ $$\lambda = \frac{2 \pm \sqrt{44}}{2}$$

Simplify $\sqrt{44}$:

$$\sqrt{44} = \sqrt{4 \times 11} = \sqrt{4} \times \sqrt{11} = 2\sqrt{11}$$

Substitute this back into the expression for $\lambda$:

$$\lambda = \frac{2 \pm 2\sqrt{11}}{2}$$

Factor out 2 from the numerator:

$$\lambda = \frac{2(1 \pm \sqrt{11})}{2}$$ $$\lambda = 1 \pm \sqrt{11}$$

The values of $\lambda$ are $1 + \sqrt{11}$ and $1 - \sqrt{11}$.

The final answer is $1 \pm \sqrt{11}$.