(1.) We let X be the width. So we write: L = 5 + W. We write the area as: 19 = (5 + W)(W) --> 0 = W^2 + 5W - 19 --> We can't solve the equation by factoring, so we solve this by quadratic formula. So we get: W = -7.52, 2.52. We can't have a negative width so we use 2.52: L = 2.52 + 5 = 7.52. Length: 7.52 m long. (2.) We let X be the 1st number and X + 1 be the 2nd number. We write: (X)(X + 1) = 650 --> X^2 + X - 650 = 0 --> (X - 25)(X + 26) = 0 --> X = -26, 25. So the numbers are: (25 and 26) and (-26 and -25). (3.) We use completing the square to find C: 3[x² - (8/3)x + (8/6)^2 = 0] --> 3[x² - (8/3)x + (64/36) = 0] --> 3x² - 8x + (32/9) = 0 (4.) Let X be the original length of the lot. We write: (X)(X + 8) = 345 --> X^2 + 8X - 345 = 0 --> (X - 15)(X + 23) = 0 --> We use X = 15. The new dimensions are 15 m by 23 m.