Question 195150


{{{0=-16t^2+5t+15}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{0=at^2+bt+c}}} where {{{a=-16}}}, {{{b=5}}}, and {{{c=15}}}



Let's use the quadratic formula to solve for t



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



{{{t = (-(5) +- sqrt( (5)^2-4(-16)(15) ))/(2(-16))}}} Plug in  {{{a=-16}}}, {{{b=5}}}, and {{{c=15}}}



{{{t = (-5 +- sqrt( 25-4(-16)(15) ))/(2(-16))}}} Square {{{5}}} to get {{{25}}}. 



{{{t = (-5 +- sqrt( 25--960 ))/(2(-16))}}} Multiply {{{4(-16)(15)}}} to get {{{-960}}}



{{{t = (-5 +- sqrt( 25+960 ))/(2(-16))}}} Rewrite {{{sqrt(25--960)}}} as {{{sqrt(25+960)}}}



{{{t = (-5 +- sqrt( 985 ))/(2(-16))}}} Add {{{25}}} to {{{960}}} to get {{{985}}}



{{{t = (-5 +- sqrt( 985 ))/(-32)}}} Multiply {{{2}}} and {{{-16}}} to get {{{-32}}}. 



{{{t = (-5+sqrt(985))/(-32)}}} or {{{t = (-5-sqrt(985))/(-32)}}} Break up the expression.  



{{{t = (5-sqrt(985))/(32)}}} or {{{t = (5+sqrt(985))/(32)}}} Reduce



So the answers are {{{t = (5-sqrt(985))/(32)}}} or {{{t = (5+sqrt(985))/(32)}}} 



which approximate to {{{t=-0.825}}} or {{{t=1.137}}}