How do I put this in the quadratic form: vt -1/2gt^2 in terms of t?

Question

It is supposed to look like t= v/g +- (v^2/g^2 -2y/g)^1/2

my brain hurts :- {

Answer 1

y=vt-1/2gt^2

-1/2gt^2+vt-y=0

Use the Quadratic Formula

t={-b+/-rt(b^2-4ac)} / 2a , where

a=co-efficient of t^2, = -1/2g

b=co-efficient of t, =v

c= constant term, = -y

t={-v+/- rt(v^2-4(-1/2)(g)(-y)} / 2(-1/2)g

t={-v+/-rt(v^2-2gy)} /-g

That's the answer. From here on, it's a matter of jiggling terms to make

it look like the mess you described.

t={-v+/-rt(v^2-2gy)} / -g

t= -v/-g +/-rt(v^2-2gy)/ -g

t=v/g +/- rt[v^2/g^2 -2gy/g^2] *Note that I put the -g denominator back

under the square root sign by making it g^2. However, in doing this

my -g, when squared, became +g^2 Not a problem.

t=v/g+/-rt(v^2/g^2-2y/g)

Since the exponent symbol for "root" is ^1/2, we get

t=v/g+/-(v^2/g^2-2y/g)^1/2

My brain hurts now too!!!

Cheers