The given equation is:

$1 + 2 + 3 + 4 + \dots + x = \frac{(n-8)(n-7)}{2}$

This represents the sum of the first $x$ natural numbers on the left-hand side. The sum of the first $x$ natural numbers is given by:

$\frac{x(x+1)}{2}$

Thus, we can set this equal to the right-hand side of the given equation:

$\frac{x(x+1)}{2} = \frac{(n-8)(n-7)}{2}$

We can eliminate the $\frac{1}{2}$ from both sides:

$x(x+1) = (n-8)(n-7)$

Now, to express $x$ in terms of $n$, we'll solve this quadratic equation:

$x^2 + x = (n-8)(n-7)$

Expanding the right-hand side:

$x^2 + x = n^2 - 15n + 56$

Rearranging the equation:

$x^2 + x - (n^2 - 15n + 56) = 0$

Thus, the quadratic equation in $x$ becomes:

$x^2 + x - n^2 + 15n - 56 = 0$

Now, solve this quadratic equation using the quadratic formula:

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-n^2 + 15n - 56)}}{2(1)}$

Simplifying inside the square root:

$x = \frac{-1 \pm \sqrt{1 + 4(n^2 - 15n + 56)}}{2}$ $x = \frac{-1 \pm \sqrt{1 + 4n^2 - 60n + 224}}{2}$ $x = \frac{-1 \pm \sqrt{4n^2 - 60n + 225}}{2}$

Since $\sqrt{4n^2 - 60n + 225} = 2n - 15$, the expression for $x$ becomes:

$x = \frac{-1 \pm (2n - 15)}{2}$

Thus, we have two possible solutions for $x$:

$x = \frac{-1 + (2n - 15)}{2} \quad \text{or} \quad x = \frac{-1 - (2n - 15)}{2}$

Simplifying these:

$x = \frac{2n - 16}{2} = n - 8 \quad \text{or} \quad x = \frac{-2n + 14}{2} = -(n - 7)$

Since $x$ must be a positive integer, we take the solution:

$x = n - 8$

Final Answer:

$x = n - 8$

Would you like more details or have any questions about this?

Here are 5 related questions you might find interesting:

1. How do you derive the sum of the first $n$ natural numbers formula?
2. How can we generalize this method to other types of series?
3. What is the significance of solving quadratic equations in this context?
4. Can you use completing the square to solve the quadratic instead of the quadratic formula?
5. What are the conditions for a quadratic equation to have real roots?

Tip: Always simplify an equation as much as possible before solving for a variable.