Quadratic formula substitution?

[Quinch]

Or... something like that. Originating from an accidentally misfiled thread, I'm trying to find the t in this equation;

 

[math]

s = {v_0}t + \frac{1}{2}at^2

[/math]

 

Normally I'd try to shuffle the values around, adding, subtracting, dividing etc. until I ended up with what I wanted, but I've long since forgotten how to deal with multiple exponents and barely remembering the terminology {let alone in English} didn't help finding an answer either.

 

Would someone be so kind as to give me a pointer or two how to deal with this?

[ajb]

Use The Formula.

[Klaynos]

Do what ajb said, you'll need to rearrange what you've got though.

[Air]

Or... something like that. Originating from an accidentally misfiled thread, I'm trying to find the t in this equation;

 

[math]

s = {v_0}t + \frac{1}{2}at^2

[/math]

 

Normally I'd try to shuffle the values around, adding, subtracting, dividing etc. until I ended up with what I wanted, but I've long since forgotten how to deal with multiple exponents and barely remembering the terminology {let alone in English} didn't help finding an answer either.

 

Would someone be so kind as to give me a pointer or two how to deal with this?

 

When you rearrange the formula, you are able to get it into the form [math]ax^2 + bx + c =0[/math]. [math]s = {v_0}t + \frac{1}{2}at^2 \rightarrow \frac12 at^2 + {v_0}t - s = 0[/math]. This implies:

 

[math]a = \left(\frac12 a\right), \ b={v_0}, \ c={-s}[/math].

 

 

For these type of formula, known as the quadratic equation (Highest power is 2), you can use the quadratic equation to find the value of [math]x[/math]. The quadratic formula is:

 

[math]x= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math]