Answer
The width is $(-5+5\sqrt{17})$ $ft$ and the length is $(5+5\sqrt{17})$ $ft$
Work Step by Step
Let $x$ be the width of the rectangle. Hence, $x+10$ is the length.
Using $A=lw$ or the formula for the area of a rectangle, then
\begin{array}{l}\require{cancel}
400=x(x+10)
.\end{array}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
400=x^2+10x
\\
0=x^2+10x-400
.\end{array}
In the equation above, $a=
1
,$ $b=
10
,$ and $c=
-400
.$ Using the Quadratic Formula which is given by $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
,$ then
\begin{array}{l}\require{cancel}
x=\dfrac{-10\pm\sqrt{10^2-4(1)(-400)}}{2(1)}
\\
x=\dfrac{-10\pm\sqrt{100+1600}}{2}
\\
x=\dfrac{-10\pm\sqrt{1700}}{2}
\\
x=\dfrac{-10\pm\sqrt{100\cdot17}}{2}
\\
x=\dfrac{-10\pm\sqrt{10^2\cdot17}}{2}
\\
x=\dfrac{-10\pm10\sqrt{17}}{2}
\\
x=\dfrac{\cancel2(-5)\pm\cancel2(5)\sqrt{17}}{\cancel2}
\\
x=-5\pm5\sqrt{17}
.\end{array}
Since $x$ is a measurement (and hence should be nonnegative), then $
x=-5+5\sqrt{17}
.$
Hence, the width, $x,$ is $(-5+5\sqrt{17})$ $ft.$ The length, $x+10$ is $(5+5\sqrt{17})$ $ft.$
