Answer:

The actual dimension is 18.28 by 3.28
Step-by-step explanation:
Given
Dimension:


Required
Determine the quadratic equation and get the possible values of x
Solving (a): Quadratic Equation.
The cardboard is rectangular in shape.
Hence, Area is calculated as thus:


Open Bracket

Subtract 60 from both sides

<em>Hence, the above represents the quadratic equation</em>
Solving (b): The actual dimension
First, we need to solve for x
This can be solved using quadratic formula:

Where



So:





Split:
or 
or 
or 
But length can't be negative;
So:

The actual dimensions:
is




<em>The actual dimension is 18.28 by 3.28</em>