Let
x be the width
x + 25 be the length
Given that the area is 7154 square feet, then we can substitute the following using the formula for the area of a rectangle
[tex]\begin{gathered} A_{\text{rectangle}}=lw \\ 7154=(x+25)(x) \end{gathered}[/tex]
Simplify the left side and equate to zero
[tex]\begin{gathered} 7154=(x+25)(x) \\ 7154=x^2+25x \\ x^2+25x-7154=0 \end{gathered}[/tex]
Now it is in the standard quadratic equation with the following coefficients, a = 1, b = 25, and c = -7154.
Use the quadratic formula to solve for x
[tex]\begin{gathered} x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \\ x = \frac{ -25 \pm \sqrt{25^2 - 4(1)(-7154)}}{ 2(1) } \\ x = \frac{ -25 \pm \sqrt{625 - -28616}}{ 2 } \\ x=\frac{-25\pm\sqrt{625+28616}}{2} \\ x = \frac{ -25 \pm \sqrt{29241}}{ 2 } \\ x = \frac{ -25 \pm 171\, }{ 2 } \\ \\ x_1=\frac{-25+171}{2} \\ x_1=\frac{146}{2} \\ x_1=73 \\ \\ x_2=\frac{-25-171}{2} \\ x_2=\frac{-196}{2} \\ x_2=-98 \end{gathered}[/tex]
There are two solutions for x, we will be discarding x = -98, as there is no negative dimensions.
With x = 73, we have the following
[tex]\begin{gathered} \text{width }=73 \\ \text{length }=73+25=98 \end{gathered}[/tex]
Therefore, the dimensions are 73 feet for width, and 98 feet for length.
