How would you solve an equation like this?

Multiply by both sides by a constant so that all of the coefficients are integers, then solve it like any other quadratic equation.

quack wrote:

Multiply by both sides by a constant so that all of the coefficients are integers, then solve it like any other quadratic equation.

Don't bother multiplying both sides by a constant. There is no need for integers.

Solve it like any other quadratic equation.

Or just solve it as is using the quadratic equation.

metarded wrote:0.48 + 3.55t -4.9(t^2) = 0, solve for t

ax^2 + bx + c = 0

x = [-b +- sqrt(b^2 - 4ac)]/(2a)

Your equation is obviously a physics related one, so ask yourself which one of the answers (there will be two) makes the most sense in regard to the original problem.

Try Wolfram|Alpha.

http://www.wolframalpha.com/input/?i=solve%280.48+%2B+3.55

*t+-4.9*t^2+%3D+0%2C+t%29

t = {-0.11648315612886409, 0.8409729520472315}

back in college i knew how to solve those equations. like most people i thought to myself "when in real life and i going to need to know this?" so far so good!

It's a quadratic. Find the roots and solve accordingly. If you don't know how to do that then nightschool is in order.

Go buy a TI-89 and enter "solve(0.48+3.55t-4.9(t^2)=0,t)

Hey I didn't have time to read the whole thread, has anyone suggested the quadratic equation yet?

Festizio wrote:

back in college i knew how to solve those equations. like most people i thought to myself "when in real life and i going to need to know this?" so far so good!

life is generally pretty simple when your job only involves transferring things from a deep fryer to some wax paper.

big finance guy wrote:

life is generally pretty simple when your job only involves transferring things from a deep fryer to some wax paper.

Or wasting your time on letsrun...

I would plug in 1 and realize the answer is just

a bit less than that. Or maybe I would get graph paper and

plot the answers for t = -1, 0, 1, then just extrapolate.

Or maybe I'd move the .48 over to the right side and then solve giving myself 1' to get three decimal places. Or maybe I'd let my mind wander to the Wolfram lecture on how math education needs to change.

If stuck at an airport, I'd try to turn this into a story

problem about a midget doing box jumps into a headwind.

you need to find the eigenvectors, bro. It's all about the eigenvectors

that's "wectors" to you, bro.

Well, let's see...

First you need to multiply both sides of the equation by e^(-t):

0.48*e^(-t) + 3.55*t*e^(-t) - 4.9*t^2*e^(-t) = 0;

Then integrate from 0 to infinity and you get:

-5.77

Multiply by e^(pi*i) to account for crosswinds and divide by 7 to account for steric hindrance.

0.82428

There are 59.94 interlaced fields per second for NTSC systems, so add the reciprocal of this to your value and you get:

0.84097

Voilà! This is the correct value of t. Go ahead and test it if you wish.

it has ugly coefficients which are off-putting

just put them into here :

http://zonalandeducation.com/mmts/miscellaneousMath/quadraticRealSolver/quadraticRealSolver.html

t = 0.8410 & -0.1165

like someone said, this looks like a physics problem, so your required answer is the 1st one

Taylor Series

You'll need the following items:

1. A meter stick

2. An atomic clock

3. The ability to throw a ball at precisely 3.55 m/s

Once you have those let me know and I'll describe the next steps.