Question 349686
Let the width be w and the length be l
{{{w^2 + l^2= 15}}}} pythagorian theorem
{{{w = l+3}}}
{{{w^2 + (w+3)^2 = 15}}}
{{{w^2 + w^2 + 6w + 9 = 15}}}
{{{2w^2 + 6w - 6 = 0}}}
{{{w^2 + 3w - 3 = 0}}}
*[invoke quadratic "x", 1, 3, -3 ] 

The width cannot be nagative, so it must be 0.79
Which makes the length 3.79
Check your answer. Does
{{{(3.79)^2 + (0.79)^2 = 15}}}?