PLEASE HELP QUICK, I NEED ITSolve for t.d=−16t2+12tt=38±9−4d√8t=−12±9−4d√8t=12±9−4d√2t=38±49−4d−−−−−√
Accepted Solution
A:
We take the equation d = -16t^2+12t and subtract d from both sides to get 0 = -16t^2+12t - d We apply the quadratic formula to solve for t. With a = -16, b = 12, c = -d, we have t = [ -(12) ± √( 12^2 - 4(-16)(-d) ) ] / [2 * -16] = [- 12 ± √(144-64d) ] / (-32) = [- 12 ± √16(9-4d)] / (-32) = [- 12 ± 4√(9-4d)] / (-32) = 3/8 ±√(9-4d) / 8 The answer to your question is t = 3/8 ±√(9-4d) / 8