Let the length of the rectangle be $L$ (in units).

Since the width is 6 units less than the length, we can express the width as: $W = L - 6$

The area of a rectangle is given by: $\text{Area} = L \times W$

We know the area is 7 square units, so: $L \times (L - 6) = 7$

Expanding this equation: $L^2 - 6L = 7$

Rearrange it to form a standard quadratic equation: $L^2 - 6L - 7 = 0$

Now, we can solve this quadratic equation using the quadratic formula: $L = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}$ $L = \frac{6 \pm \sqrt{36 + 28}}{2}$ $L = \frac{6 \pm \sqrt{64}}{2}$ $L = \frac{6 \pm 8}{2}$

This gives two possible solutions for $L$:

1. $L = \frac{6 + 8}{2} = 7$
2. $L = \frac{6 - 8}{2} = -1$

Since the length cannot be negative, we discard $L = -1$.

Thus, the length of the rectangle is: $\boxed{7 \text{ units}}$

Would you like further details, or do you have any questions about the solution?

Here are some related questions to consider:

1. What would the width of the rectangle be if the length is 7 units?
2. How would the area change if the length were doubled?
3. What if the area was given as a different value, like 10 square units—how would you set up the equation?
4. How can you check if the dimensions found satisfy both the area and width requirements?
5. Can you apply similar steps to solve for a square’s side given its area?

Tip: When dealing with quadratic equations, always check if both solutions make sense in the context of the problem.