simple algebra question! 10 pts to best answer?

This is a simple physics equation for position of a particle subject to an initial velocity and an acceleration. It is a second order equation, so simply use the quadratic formula. First, put it into simple form:

0=0.5at^2+vt-x

t1 & t2 = {-b +-SQRT[v^2-4(0.5a)(-x)] } / 2(0.5a)

Astrobuf

This seems like a Physics equation for constant acceleration!

Well, this is a quadratic equation which you can rewrite as 0 = (1/2)at^2 + vt - x.

So, a = 1/2

b = v

c = -x

Use the quad. eqtn. x = [-b +/- sqrt(b^2 - 4ac)] / 2a

So, t = [-v +/- sqrt(v^2-4(1/2)(-x))] / 2(1/2)

t = -v +/- sqrt(v^2 + 2x)

[but t, of course, should always be positive]

cori merci is puzzled. have faith the 1st and nil.33 answer. even nonetheless the question might desire to be mis interpreted (-3)squared ability (-3)(-3) = 9 -3squared might desire to characterize -(3)(3) = -9 however the challenge is maximum possibly 9 the unfavourable time unfavourable=unfavourable is incorrect x - 3^2 IS x -9 (-3)^2 is 9 -3^2 is -9 this could be a challenge with typing arithmetic.....

if its so simple your get your answer :P