The length of a photograph is 5 cm greater than its width. A margin of uniform width is placed around the photograph. If the total area of the margin is 300 cm2 and the dimensions of the frame (photog

The equation is $2x^2 - 51x + 115 = 0$.
We need two numbers that multiply to $2 \times 115 = 230$ and add up to $-51$.
These numbers are $-5$ and $-46$.
Rewrite the middle term:

The width of the margin $x$ must be less than half of the smallest dimension of the frame. The frame width is 21 cm, so $2x$ must be less than 21 cm, meaning $x < 10.5$ cm.
If $x = 23$ cm, then $2x = 46$ cm. This would make the picture width $21 - 46 = -25$ cm, which is not possible. So $x = 23$ is not a valid solution.
If $x = 2.5$ cm, then $2x = 5$ cm. This is a valid margin width.

Picture length = $30 - 2x = 30 - 2(2.5) = 30 - 5 = 25$ cm
Picture width = $21 - 2x = 21 - 2(2.5) = 21 - 5 = 16$ cm

The dimensions of the picture are 25 cm by 16 cm.
The final answer is $\text{Length} = 25 \text{ cm, Width} = 16 \text{ cm}$.