The problem asks to find the value(s) of $x$ given that a rectangle $PQRS$ has two squares of side $x$ cut from it. The rectangle has sides of length $20$ cm and $10+x+10 = 20+x$ cm. The area of the shaded region is $484$ cm$^2$.

1. Problem Description

The problem asks to find the value(s) of $x$ given that a rectangle $PQRS$ has two squares of side $x$ cut from it. The rectangle has sides of length $20$ cm and $10+x+10 = 20+x$ cm. The area of the shaded region is $484$ cm$^2$.

2. Solution Steps

The area of rectangle $PQRS$ is the product of its sides, which is $20(20+x) = 400 + 20x$.

The area of the two squares that are cut out is $x^2 + x^2 = 2x^2$.

The area of the shaded region is the area of the rectangle minus the area of the two squares, so $400 + 20x - 2x^2 = 484$.

Rearranging the equation, we have $2x^2 - 20x + 484 - 400 = 0$, which simplifies to $2x^2 - 20x + 84 = 0$.

Dividing by 2 gives $x^2 - 10x + 42 = 0$.

We can solve the quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-10$, and $c=42$.

$x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(42)}}{2(1)}$

$x = \frac{10 \pm \sqrt{100 - 168}}{2}$

$x = \frac{10 \pm \sqrt{-68}}{2}$

Since the discriminant is negative, there are no real solutions for $x$.

Let's re-examine the original problem. The rectangle $PQRS$ has sides of lengths $20$ cm and $20+x$ cm. Two squares with sides of length $x$ cm are cut. The area of the shaded region is $484$ cm$^2$. So the area of the rectangle minus the area of the two squares should equal the area of the shaded region.

The area of rectangle $PQRS$ is $20 \times (20+x) = 400 + 20x$.

The total area of the two squares is $2x^2$.

Thus, $400 + 20x - 2x^2 = 484$.

Rearranging gives $2x^2 - 20x + 84 = 0$.

Dividing by 2 gives $x^2 - 10x + 42 = 0$.

Since the problem implies there are real solutions, there could be a typo in the area of the shaded portion. Let's assume it should be 384 cm$^2$.

Then $400 + 20x - 2x^2 = 384$, so $2x^2 - 20x - 16 = 0$.

Dividing by 2 gives $x^2 - 10x - 8 = 0$.

Using the quadratic formula,

$x = \frac{10 \pm \sqrt{100 - 4(1)(-8)}}{2} = \frac{10 \pm \sqrt{132}}{2} = \frac{10 \pm 2\sqrt{33}}{2} = 5 \pm \sqrt{33}$.

Since $x$ must be positive, we take the positive root: $x = 5 + \sqrt{33} \approx 5 + 5.74 \approx 10.74$.

Alternatively, assuming the area is 484, we should look for mistakes made transcribing or understanding the problem. If instead of two squares, only one square was cut out, then we would have $400+20x - x^2 = 484$, which gives $x^2 - 20x + 84 = 0$. Then $x = \frac{20 \pm \sqrt{400-4(84)}}{2} = \frac{20 \pm \sqrt{64}}{2} = \frac{20 \pm 8}{2}$. So $x = 14$ or $x = 6$.

If $x=14$, then the width of the rectangle is $20+x=34$, and the area of the rectangle is $20 \times 34 = 680$. Then $680 - 14^2 = 680 - 196 = 484$. This solution works for one square.

Since the image indicates TWO squares are cut from the rectangle, the previous reasoning is correct.

However, because the discriminant is negative, there is no real solution. There is an error in the statement of the problem or diagram.

3. Final Answer

There are no real values of $x$ that satisfy the given conditions.