Question 668808
I'm assuming you have the equation {{{-16t^2+22t+3 = 0}}}


Use the quadratic formula to solve for t


{{{t = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{t = (-(22)+-sqrt((22)^2-4(-16)(3)))/(2(-16))}}} Plug in {{{a = -16}}}, {{{b = 22}}}, {{{c = 3}}}


{{{t = (-22+-sqrt(484-(-192)))/(-32)}}}


{{{t = (-22+-sqrt(484+192))/(-32)}}}


{{{t = (-22+-sqrt(676))/-32}}}


{{{t = (-22+sqrt(676))/-32}}} or {{{t = (-22-sqrt(676))/-32}}}


{{{t = (-22+26)/-32}}} or {{{t = (-22-26)/-32}}}


{{{t = 4/-32}}} or {{{t = -48/-32}}}


{{{t = -1/8}}}    or    {{{t = 3/2}}}


So the two solutions are {{{t = -1/8}}}    or    {{{t = 3/2}}}