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The length of the diagonal of a rectangle is 3 more than the width and the width is 9 less than the length if the area is 580 find the length of the rectangle

Algebra 1
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Then according to the problem, the length will be X, The width will be X-9; the length of the diagonal is (X-6). The area is equal to length times width = 580 =  X*(X-9)= X^2 - 9x, readjusting the equation we obtain
X^2 - 9X -580 =0; solving this quadratic and using the positive root, the result is that the length value comes to be 29, the width will be 20.  Checking Pythagoras we have that 23^2 = 20^2 + 29^2. and will give that is more than the 20+3 given in the problem. Check the data.