Question 189550


{{{0=-16t^2+8t+4}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{at^2+bt+c}}} where {{{a=-16}}}, {{{b=8}}}, and {{{c=4}}}



Let's use the quadratic formula to solve for t



{{{t = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{t = (-(8) +- sqrt( (8)^2-4(-16)(4) ))/(2(-16))}}} Plug in  {{{a=-16}}}, {{{b=8}}}, and {{{c=4}}}



{{{t = (-8 +- sqrt( 64-4(-16)(4) ))/(2(-16))}}} Square {{{8}}} to get {{{64}}}. 



{{{t = (-8 +- sqrt( 64--256 ))/(2(-16))}}} Multiply {{{4(-16)(4)}}} to get {{{-256}}}



{{{t = (-8 +- sqrt( 64+256 ))/(2(-16))}}} Rewrite {{{sqrt(64--256)}}} as {{{sqrt(64+256)}}}



{{{t = (-8 +- sqrt( 320 ))/(2(-16))}}} Add {{{64}}} to {{{256}}} to get {{{320}}}



{{{t = (-8 +- sqrt( 320 ))/(-32)}}} Multiply {{{2}}} and {{{-16}}} to get {{{-32}}}. 



{{{t = (-8 +- 8*sqrt(5))/(-32)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{t = (-8+8*sqrt(5))/(-32)}}} or {{{t = (-8-8*sqrt(5))/(-32)}}} Break up the expression.  



{{{t = (1-sqrt(5))/(4)}}} or {{{t = (1+sqrt(5))/(-32)}}} Reduce



So the answers are {{{t = (1-sqrt(5))/(4)}}} or {{{t = (1+sqrt(5))/(-32)}}} 



which approximate to {{{t=-0.309}}} or {{{t=0.809}}} 



Since a negative time doesn't make sense, this means that the only logical solution is {{{t=0.809}}}



So it takes about 0.809 seconds for the object to hit the ground