Question 30649
LEngth = x+5
Width = x
Equation:
*[tex (x+5)(x)=300]

*[tex x^2+5x-300=0]
<pre>Solve for x
a=1, b=5, c=-300
{{{x=(-5+-sqrt(5^2-4*1*-300))/(2*1)}}}
{{{x=(-5+-sqrt(25+1200))/(2*1)}}}
{{{x=(-5-35)/2}}} or {{{x=(-5+35)/2}}}
Remove negative:
x=15
15+5=20


<font color = green>Hence, the length is 20ft and the width is 15ft.</font>Paul.