# Solving for a variable with quadratic equations

• Oct 17th 2010, 07:04 PM
Kyrie
Solving for a variable with quadratic equations
Solving for t

$\displaystyle A = 6.5 - \frac{20.4t}{t^2+36}$

I wasn't sure but I decided to try to get the fraction out by multiplying both sides by $\displaystyle t^2+36$

Which makes it:

$\displaystyle A(t^2+36) = 6.5(t^2+36) - 20.4t$

I then distributed accordingly.

$\displaystyle At^2 + A36 = 6.5t^2 + 234 - 20.4t$

I don't know if I'm doing this correctly or where to go next. Help appreciated immensely. <3
• Oct 17th 2010, 07:10 PM
Educated
You are on the right track.

Move everything to one side and factorise to get the formula to look similar to: $\displaystyle (a)t^2 + (b)t + (c) = 0$
• Oct 17th 2010, 07:11 PM
pickslides
Move everything to one side.

$\displaystyle \displaystyle At^2 + A36 = 6.5t^2 + 234 - 20.4t$

$\displaystyle \displaystyle At^2 + A36 - 6.5t^2 - 234 + 20.4t= 0$

now group in order of $\displaystyle t$

$\displaystyle \displaystyle At^2 - 6.5t^2 + 20.4t+ A36- 234= 0$

Factoring

$\displaystyle \displaystyle (A - 6.5)t^2 + 20.4t+ A36- 234= 0$

Now in the quadratic formula make $\displaystyle \displaystyle a = A - 6.5 , b =20.4, c= A36- 234= 0$
• Oct 18th 2010, 12:04 AM
Kyrie
Thanks guys.

I get to here:
$\displaystyle \frac{-20.4 ± \sqrt{20.4^2 - 4 (A-6.5)(A36-234)}}{2 (A-6.5)}$

Do I foil and multiply out? This problem frustrates me. (Thinking)
• Oct 18th 2010, 01:59 AM
HallsofIvy
Not unless you have so good reason to do so. That is the answer.