To solve the problem, we need to find the dimensions of the rectangular poster. Let's denote the width of the poster as $w$ inches. Since the length is 9 inches longer than the width, the length can be expressed as $w + 9$ inches.

Given that the diagonal of the rectangle is 15 inches, we can use the Pythagorean theorem, which states that:

$\text{Diagonal}^2 = \text{Length}^2 + \text{Width}^2$

Substituting the given values into the equation:

$15^2 = (w + 9)^2 + w^2$

Now, let's solve for $w$.

$225 = (w + 9)^2 + w^2$

Expand the square:

$225 = w^2 + 18w + 81 + w^2$

Combine like terms:

$225 = 2w^2 + 18w + 81$

Subtract 225 from both sides to set the equation to zero:

$0 = 2w^2 + 18w + 81 - 225$

$0 = 2w^2 + 18w - 144$

Divide the entire equation by 2 to simplify:

$0 = w^2 + 9w - 72$

Now, we solve this quadratic equation using the quadratic formula:

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

For our equation, $a = 1$, $b = 9$, and $c = -72$:

$w = \frac{-9 \pm \sqrt{9^2 + 4 \times 72}}{2}$

$w = \frac{-9 \pm \sqrt{81 + 288}}{2}$

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