I'm using an intermediate algebra textbook and it had this problem:

"Solve the formula $$h = 48t + \frac{1}{2}a t^2$$ for $t$." The answer they displayed was: $$a = \frac{2h-96t}{t^2}$$

Can anyone tell me how this answers the question "solve for $t$", and what would be the right way to solve for $t$?

  • 1
    $\begingroup$ That is solving for a. $\endgroup$ – Ian May 18 '19 at 3:53
  • $\begingroup$ Maybe they meant solve for a. $\endgroup$ – NoLand'sMan May 18 '19 at 3:54
  • $\begingroup$ So, what would the answer for solving for t? $\endgroup$ – Joe May 18 '19 at 4:05

If you want to solve for $a$, you want to isolate $a$ to get it by itself.

So first, I'd subtract $48t$ from both sides.

$$h - 48t = \frac{1}{2}at^2$$

Then you can multiply both sides by 2:

$$ 2(h-48t) = at^2$$

and finally dividing both sides by $t^2$

$$a =\frac{2h-96t}{t^2}$$

Edit: to solve for $t$, you can use the quadratic formula. For $Ax^2+Bx+C = 0$, we have that

$$x = \frac{-B \pm \sqrt{B^2-4AC}}{2A}$$

So if we have $\frac{1}{2}at^2+48t-h = 0$, we can plug into the equation with $A = \frac{1}{2}a$, $B = 48, C=-h$:

$$t = \frac{-48 \pm \sqrt{48^2 - 4(\frac{1}{2}a)(-h)}}{2(\frac{1}{2}a)}$$

  • $\begingroup$ What would be the answer for solving for t? $\endgroup$ – Joe May 18 '19 at 4:05
  • $\begingroup$ @user65141 see my edit $\endgroup$ – rb612 May 18 '19 at 4:10

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