The flower garden has the shape of a right triangle. 68 ft of a perennial border forms the hypotenuse of the triangle, and one leg is 28 ft longer than other leg. find the lengths of the legs.

Let x ft be the dimension of one leg. Then, the other leg will be (x+28) ft.
Applying Pythagoras theory;
Hypotenuse = Sqrt [x^2 +(x+28)^2]

68 = Sqrt [x^2 + x^2+28x+28x+784]
68^2 = 2x^2+56x+784
2x^2 + 56x +784 - 4624 = 0
2x^2 + 56x - 3840 = 0

Solving for x;
x = [-b +/- Sqrt (b^2 -4ac)]/2a where a = 2, b = 56, c = -3840
Then,
x = [-56+/- Sqrt (56^2-4*2*(-3840)]/2*2 = -14+/-46
x = 32 or - 60
Ignoring the negative value, x = 32 ft
Therefore,
One leg dimension = 32 ft
Second leg dimension = 32+28 = 60 ft

The answer is 16.66 i'm pretty sure