Solve for x: Rectangle Area x²-13 with Dimensions (x-4) and (x+1)

First, recall the formula for calculating the area of a rectangle with sides of length a,b (length units):

$S_{\boxed{\hspace{8pt}}}=a\cdot b$

Therefore, by direct calculation, for the rectangle shown in the drawing with side lengths:

$x+1,\hspace{6pt}x-4$ (length units),

The expression for the area is:

$$$S_{\boxed{\hspace{8pt}}}=(x+1)(x-4)$

However, from the given information, we know that the expression for the area of the rectangle in the drawing is:

$S_{\boxed{\hspace{8pt}}}=x^2-13$

Therefore, we can conclude the existence of the equation:

$(x+1)(x-4)=x^2-13$

Now, in order to simplify the equation, recall the expanded distribution law:

$(a+b)(c+d)=ac+ad+bc+bd$

Proceed to solve the equation that we obtained. First, we'll open the parentheses on the left side, then we'll move and combine like terms, and solve the resulting simple equation:

$(x+1)(x-4)=x^2-13 \\ x^2-4x+x-4=x^2-13\\ -3x=-10\hspace{9pt}\text{/:(-3)}\\ \boxed{x=3}$(length units),

Note- this solution for the unknown does not contradict the domain of definition (where the side lengths must be positive, as required) and the area obtained by substituting it into the given expression for the area in the problem:

$S_{\boxed{\hspace{8pt}}}=x^2-13 \\ \downarrow\\ S_{\boxed{\hspace{8pt}}}=5^2-13 =25-13=\boxed{12}$(area units)

Indeed positive, as expected.

Therefore, the correct answer is answer C.