It looks like you're trying to solve a problem involving the sum of an arithmetic series. Let's break it down step by step.

The sum of the first \( n \) natural numbers is given by the formula:
\[ S = \frac{n(n + 1)}{2} \]

You have two equations:
1. \( 1 + 2 + \ldots + x = 55 \)
2. \( 1 + 2 + \ldots + x = 55 \)

It seems there might be a typo or repetition in your problem statement. Let's focus on solving the first equation:

\[ \frac{x(x + 1)}{2} = 55 \]

To solve for \( x \), follow these steps:

1. Multiply both sides by 2 to get rid of the fraction:
\[ x(x + 1) = 110 \]

2. Rearrange the equation into standard quadratic form:
\[ x^2 + x - 110 = 0 \]

3. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -110 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-110)}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{1 + 440}}{2} \]
\[ x = \frac{-1 \pm \sqrt{441}}{2} \]
\[ x = \frac{-1 \pm 21}{2} \]

This gives us two potential solutions:
\[ x = \frac{20}{2} = 10 \]
\[ x = \frac{-22}{2} = -11 \]

Since \( x \) must be a positive integer, we discard \( x = -11 \). Therefore, the solution is:
\[ x = 10 \]

So, the value of \( x \) that satisfies the equation \( 1 + 2 + \ldots + x = 55 \) is \( x = 10 \).