Abridge on a river is modeled by the equation h = -0.2d^2 + 2.25d, where h is the height and d is the horizontal distance. for cleaning and maintenance purposes a worker wants to tie a taut rope on two ends of the bridge so that he can slide on the rope. the rope is at an angle defined by the equation -d + 6h = 21.77. if the rope is attached to the bridge at points a and b, such that point b is at a higher level than point a, at what distance from the ground level is point a?

Hello,
A=(2.21082279...,3.996803798)

Since you are interested in values of h, it is convenient to eliminate d from the equations. The second equation tells you
  d = 6h - 21.77

Substituting that into the first equation gives
  h = (6h -21.77)(-0.2(6h -21.77) +2.25) = (6h -21.77)(-1.2h +6.604)
  7.2h² -64.748h +143.76908 = 0 . . . . rearranging to standard form

The quadratic formula tells you the solution to
  ax²+bx+c=0
is ...
  x = (-b±√(b²-4ac))/(2a)

Using this formula on our quadratic, we have
  h = (64.748 ±√((-64.748)² -4(7.2)(143.76908)))/(2·7.2)
  h = (64.748 ±√51.754)/14.4
  h ≈ 3.997 or 4.996

Point A is 3.997 from ground level.