Question 483670
<pre>

{{{h=expr(1/2)gt^2+v[0]t}}}

As you said, multiply through by 2

{{{2h=gt^2+2v[0]t}}}

{{{0 = gt^2+2v[0]t-2h}}}

{{{gt^2+2v[0]t-2h=0}}}

That is a quadratic equation, so we use the
quadratic formula:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

with x = t, a = g, b = 2v<sub>0</sub>, and c = -2h

{{{t = (-(2v[0]) +- sqrt((2v[0])^2-4*g*(-2h) ))/(2*g) }}} 

{{{t = (-2v[0] +- sqrt( 4v[0]^2+8gh) )/(2*g) }}}

{{{t = (-2v[0] +- sqrt( 4(v[0]^2+2gh)) )/(2*g) }}}

{{{t = (-2v[0] +- 2sqrt( v[0]^2+2gh) )/(2*g) }}}

{{{t = (2(-v[0] +- sqrt( v[0]^2+2gh)) )/(2*g) }}}

{{{t = (cross(2)(-v[0] +- sqrt( v[0]^2+2gh)) )/(cross(2)*g) }}}

{{{t = (-v[0] +- sqrt( v[0]^2+2gh)) /g }}}

There are two solutions.  

{{{t = (-v[0] - sqrt( v[0]^2+2gh)) /g }}} and {{{t = (-v[0] + sqrt( v[0]^2+2gh)) /g }}}

The given equation is for the time when an object projected upward from the
ground is at a certain height h.  It is at that height h once going up and
once again coming down.  The first solution is for the time when the ball
first reaches height h going up, and the second is for the time when it
reaches that same height h coming down.   

Edwin</pre>