To solve the problem, let's denote the width of the original piece of cardboard as \( w \) inches. According to the problem, the length of the cardboard is \( w + 14 \) inches.
Since Lorene cuts out 7-inch squares from each corner to make the open-topped boxes, the dimensions of the box after cutting out the corners will be:
- Length: \( (w + 14) - 2(7) = w + 14 - 14 = w \)
- Width: \( w - 2(7) = w - 14 \)
- Height: \( 7 \) inches (the height of the box is the edge of the square cut out)
Given that the volume \( V \) of the box is calculated as follows:
\[ V = \text{length} \times \text{width} \times \text{height} \]
Substituting the calculated dimensions, we have:
\[ V = w \times (w - 14) \times 7 \]
According to the problem, the volume is given as \( 3360 \) cubic inches. Thus, we have:
\[ w \times (w - 14) \times 7 = 3360 \]
Dividing both sides by 7:
\[ w \times (w - 14) = \frac{3360}{7} = 480 \]
This simplifies to:
\[ w(w - 14) = 480 \]
Expanding it gives us:
\[ w^2 - 14w - 480 = 0 \]
Now we will solve this quadratic equation using the quadratic formula:
\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, \( a = 1 \), \( b = -14 \), and \( c = -480 \):
Calculating the discriminant:
\[ b^2 - 4ac = (-14)^2 - 4 \times 1 \times (-480) = 196 + 1920 = 2116 \]
Finding the square root:
\[ \sqrt{2116} = 46 \]
Now substituting back into the quadratic formula:
\[ w = \frac{14 \pm 46}{2} \]
Calculating the two possible values for \( w \):
- \( w = \frac{60}{2} = 30 \)
- \( w = \frac{-32}{2} = -16 \) (not feasible since width cannot be negative)
Thus, we have:
\[ w = 30 \]
Now, we can find the length of the original piece of cardboard:
\[ \text{Length} = w + 14 = 30 + 14 = 44 \]
Finally, the dimensions of the original piece of cardboard are:
\[ \text{Width} = 30 \text{ inches}, \quad \text{Length} = 44 \text{ inches} \]
Thus, the dimensions of the original piece of cardboard are:
\[ \boxed{30 \text{ inches} \text{ (width)} \text{ and } 44 \text{ inches} \text{ (length)}} \]