Question 747422
The new dimension is (2X+30)(2X+40) where x is the width of the walk. 

{{{(4x^X + 140X + 1200) - 624 = 1200}}}.  New area less the walk equals original area

{{{ (4X^2 + 140X - 624) = 0 }}}.  Subtract 1200 from both sides.

{{{ X^2 + 35X - 156 = 0 }}}. Divided both sides by 4

{{{ (X + 39) (X - 4) }}} 

If we were simply solving for x, we'd have -39 as one answer, but since it's a problem with dimensions, we discard, and accept the 4 only. The walk is 4 feet wide.

38 x 48 = 1824 or 624 more than the original 1200