Since the formula for #h(t)# is #22t-5t^2#, and here, #h(t)=15#, we can write:
#22t-5t^2=15#
#22t-5t^2-15=0#
#5t^2-22t+15=0#
Now use the quadratic formula to solve for #t#.
For any #ax^2+bx+c=0#, where #a!=0#, #x=(-b+-sqrt(b^2-4ac))/(2a)#.
Here, #x=t,a=5,b=-22,c=15#. Input:
#t=(-(-22)+-sqrt((-22)^2-4(5)(15)))/(2*5)#
#t=(22+-sqrt(484-300))/10#
#t=(22+sqrt(184))/10, (22-sqrt(184))/10#
#t=(22+2sqrt(46))/10, (22-2sqrt(46))/10#
#t=(11+sqrt(46))/5, (11-sqrt(46))/5#
#t=3.6s,0.8s#
You get two answers because the ball reaches its topmost point, and then comes down again. This point can be found, it is the #x# coordinate of the parabola of the function of #h(t)#:
graph{22x-5x^2 [-8.62, 11.38, 12.52, 22.52]}
As you can see, the topmost point of the stone will be at #2.2s#, the average of our two answers (as it should).
