Question 23797
let the length be y and let the width be x
y = 3x+2


Area = length x width
xy=56
x(3x+2)=56
{{{3x^2+2x=56}}}
{{{3x^2+2x-56=0}}} --- solve for x by using quardratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In {{{ax^2 + bx + c = 0}}} a=3, b=2, c=-56
{{{x = (-2 +- sqrt( (2)^2-4*3*-56 )))/ 2(3)}}} ---> divide by 2(3) 
{{{x = (-2 +- sqrt( 4+672))/6)}}}
{{{x = (-2 + sqrt( 676))/6)}}} ---> add the foloowing don't subtrract.
x = 4
3(4)+2 = 14
Hence, the length is 14 and the width is 4.