SOLUTION: How would I set up to figure the area of 560 sq ft with a length of 12 greater than the width's dimension. What are the dimensions?

Algebra ->  Algebra  -> Surface-area -> SOLUTION: How would I set up to figure the area of 560 sq ft with a length of 12 greater than the width's dimension. What are the dimensions?      Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   


Question 104616: How would I set up to figure the area of 560 sq ft with a length of 12 greater than the width's dimension. What are the dimensions?
Answer by scott8148(5885) About Me  (Show Source):
You can put this solution on YOUR website!
let x=width, so x+12=length ... x(x+12)=560 ... x^2+12x=560

x^2+12x-560=0