The area of a rectangle is 133 square feet. The length is 12 feet longer than the width. Find the length and width. 

Let's denote the width of the rectangle as \( w \) (in feet). Since the length is 12 feet longer than the width, we can express the length as \( l = w + 12 \).

The area of the rectangle is given by the formula:

\[
\text{Area} = \text{length} \times \text{width}
\]

Given that the area is 133 square feet, we can set up the equation:

\[
w \times (w + 12) = 133
\]

This simplifies to:

\[
w^2 + 12w - 133 = 0
\]

This is a quadratic equation. We can solve it using the quadratic formula:

\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( w^2 + 12w - 133 = 0 \), the coefficients are \( a = 1 \), \( b = 12 \), and \( c = -133 \). Substituting these values into the quadratic formula:

\[
w = \frac{-12 \pm \sqrt{12^2 + 4 \times 1 \times 133}}{2 \times 1}
\]

\[
w = \frac{-12 \pm \sqrt{144 + 532}}{2}
\]

\[
w = \frac{-12 \pm \sqrt{676}}{2}
\]

\[
w = \frac{-12 \pm 26}{2}
\]

This gives two possible solutions:

\[
w = \frac{-12 + 26}{2} = \frac{14}{2} = 7
\]

or

\[
w = \frac{-12 - 26}{2} = \frac{-38}{2} = -19
\]

Since the width cannot be negative, we discard \( w = -19 \).

So, the width \( w \) is 7 feet. The length \( l \) is:

\[
l = w + 12 = 7 + 12 = 19 \text{ feet}
\]

**Final Answer**:
- Width: 7 feet
- Length: 19 feet

Would you like any further details or explanations?

Here are 5 additional questions to explore related concepts:

1. How would you calculate the perimeter of this rectangle?
2. If the length was twice the width instead of 12 feet longer, what would the dimensions be?
3. Can you find the length and width if the area was 200 square feet with the same length-width relationship?
4. What would happen to the dimensions if the area was halved?
5. How do you solve a quadratic equation by completing the square?

**Tip:** When solving quadratic equations, always check the discriminant (\( b^2 - 4ac \)) first to determine the nature of the roots.

Learn how to find the length and width of a rectangle when given its area as 133 square feet and the length is 12 feet longer than the width. This problem involves algebraic concepts such as quadratic equations and the area formula of a rectangle. Suitable for students in Grades 7-9.

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