The graph of the equation $y = 5x  + 24$ intersects the graph of the equation $y = x^2$ at two points. The two points are a distance $\sqrt{N}$ apart. Find $N$.
1. **Find the points of intersection:**
   To find the points where the graphs of \( y = 5x + 24 \) and \( y = x^2 \) intersect, we set the equations equal to each other:
   \[
   5x + 24 = x^2
   \]
   Rearrange the equation to form a standard quadratic equation:
   \[
   x^2 - 5x - 24 = 0
   \]
   Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -5 \), and \( c = -24 \):
   \[
   x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1}
   \]
   \[
   x = \frac{5 \pm \sqrt{25 + 96}}{2}
   \]
   \[
   x = \frac{5 \pm \sqrt{121}}{2}
   \]
   \[
   x = \frac{5 \pm 11}{2}
   \]
   This gives us two solutions:
   \[
   x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{-6}{2} = -3
   \]

2. **Find the corresponding \( y \)-coordinates:**
   Substitute \( x = 8 \) and \( x = -3 \) back into either of the original equations to find the corresponding \( y \)-coordinates. Using \( y = x^2 \):
   \[
   y = 8^2 = 64 \quad \text{and} \quad y = (-3)^2 = 9
   \]
   Therefore, the points of intersection are \( (8, 64) \) and \( (-3, 9) \).

3. **Calculate the distance between the points:**
   The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
   \[
   d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
   \]
   Substituting the points \( (8, 64) \) and \( (-3, 9) \):
   \[
   d = \sqrt{(8 - (-3))^2 + (64 - 9)^2}
   \]
   \[
   d = \sqrt{(8 + 3)^2 + (64 - 9)^2}
   \]
   \[
   d = \sqrt{11^2 + 55^2}
   \]
   \[
   d = \sqrt{121 + 3025}
   \]
   \[
   d = \sqrt{3146}
   \]
   Therefore, \( \sqrt{N} = \sqrt{3146} \), which implies \( N = 3146 \).

The final answer is \( \boxed{3146} \)