The length of a rectangular poster is 9 more inches than two times its width. The area of the poster is 68 square inches. Solve for the dimensions (length and width) of the poster.

Let's denote the width of the rectangular poster as $w$ inches. According to the problem, the length $l$ of the poster is 9 more inches than two times its width. Therefore, we can express the length as:

$l = 2w + 9$

The area of the poster is given as 68 square inches. The area of a rectangle is calculated by multiplying its length and width:

$\text{Area} = l \times w$

Substituting the given area and the expression for the length, we get:

$68 = (2w + 9) \times w$

Now, we need to solve this equation for $w$. First, expand the right-hand side:

$68 = 2w^2 + 9w$

Next, rearrange the equation to set it to zero:

$2w^2 + 9w - 68 = 0$

This is a quadratic equation in the form $ax^2 + bx + c = 0$. We can solve it using the quadratic formula:

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

Here, $a = 2$, $b = 9$, and $c = -68$. Plugging these values into the quadratic formula, we get:

$w = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 2 \cdot (-68)}}{2 \cdot 2}$

$w = \frac{-9 \pm \sqrt{81 + 544}}{4}$

$w = \frac{-9 \pm \sqrt{625}}{4}$

$w = \frac{-9 \pm 25}{4}$

This gives us two possible solutions for $w$:

$w = \frac{-9 + 25}{4} = \frac{16}{4} = 4$

$w = \frac{-9 - 25}{4} = \frac{-34}{4} = -8.5$

Since a width cannot be negative, we discard $w = -8.5$. Therefore, the width of the poster is:

$w = 4$

Now, we can find the length using the expression $l = 2w + 9$:

$l = 2(4) + 9 = 8 + 9 = 17$

So, the dimensions of the poster are:

$\text{Width} = 4 \text{ inches}$ $\text{Length} = 17 \text{ inches}$