Quick, i need it solve for t. d=−16t2+12t t=38±9−4d√8 t=−12±9−4d√8 t=12±9−4d√2 t=38±49−4d−−−−−√

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

Here are a bunch of CORRECT answers. Your answer is in the last pic. I got number 3 wrong, but it still showed the correct answer.

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