To understand how the expression for the area of $\triangle ABC$ was simplified step by step, we start from the general formula for the area of a triangle given by Heron's formula:

$\text{Area of } \triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$

where $s$ is the semi-perimeter:

$s = \frac{a + b + c}{2}.$

Here, $a$, $b$, and $c$ are the lengths of the sides of the triangle.

Step-by-Step Simplification:

1. Substitute the side lengths into the formula:

   The semi-perimeter is $s = \frac{a + b + c}{2}$. Substituting $a = 14+x$, $b = 6-x$, and $c = 8+x$:

   $s = \frac{(14+x) + (6-x) + (8+x)}{2} = \frac{28 + x}{2}.$

   So, $s = 14 + \frac{x}{2}$.

2. Write out the terms $(s-a)$, $(s-b)$, and $(s-c)$:

   Using $s = 14 + \frac{x}{2}$:

   • $s - a = 14 + \frac{x}{2} - (14 + x) = -\frac{x}{2}$,
   • $s - b = 14 + \frac{x}{2} - (6 - x) = 8 + \frac{3x}{2}$,
   • $s - c = 14 + \frac{x}{2} - (8 + x) = 6 - \frac{x}{2}$.

3. Substitute these values into the formula:

   Now the area becomes: $\text{Area of } \triangle ABC = \sqrt{\left(14 + \frac{x}{2}\right)\left(-\frac{x}{2}\right)\left(8 + \frac{3x}{2}\right)\left(6 - \frac{x}{2}\right)}.$

4. Simplify the expressions inside the square root:

   Combine and simplify each term:

   • $s = 14 + \frac{x}{2}$,
   • $s - a = -\frac{x}{2}$,
   • $s - b = 8 + \frac{3x}{2}$,
   • $s - c = 6 - \frac{x}{2}$.

   Multiplying these together gives: $\text{Area of } \triangle ABC = \sqrt{\left(14 + x\right)(-x)(14+x)(14+x-14)(14+x-(6-x))(14+x-(8+x))}.$

Would you like me to continue this calculation step-by-step with expanded terms? Or go directly to the $area result of \[\[ \text{Area of } \triangle ABC = 4\sqrt{3}(14x + x^2).$

Would you like me to detail how to reach this simplified result?