Area of a rectangle is length by width, width is x, length is x+8
x×(x+8)=240
x²+8x=240
x²+8x-240=0
Use the quadratic formula to solve
[tex]x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}
x=\frac{-8±\sqrt{8^{2}-4(1)(-240)}}{2(1)}
x=\frac{-8±\sqrt{64-4(-240)}}{2} [/tex][tex]x=\frac{-8±\sqrt{1024}}{2}
x=\frac{-8±32}{2}
x=\frac{24}{2} OR \frac{-40}{2}
x=12 OR -20[/tex]
It is a distance so can't be negative, so the answer must be 12
