# Solving for $t$ in $h = 48t + \frac{1}{2}at^2$

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 $$t$$?

• That is solving for a. – Ian May 18 '19 at 3:53
• Maybe they meant solve for a. – NoLand'sMan May 18 '19 at 3:54
• So, what would the answer for solving for t? – 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 $$A = \frac{1}{2}a$$, $$B = 48, C=-h$$:

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

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