Question 90368
{{{80 = -16t^2 + 20t + 100}}} Start with the given equation


{{{0 = -16t^2 + 20t + 20}}} Subtract 80 from both sides


{{{0 = -16t^2 + 20t + 20}}} Subtract 80 from both sides



Let's use the quadratic formula to solve for t:



Starting with the general quadratic


{{{at^2+bt+c=0}}}


the general solution using the quadratic equation is:


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


So lets solve {{{-16*t^2+20*t+20=0}}} ( notice {{{a=-16}}}, {{{b=20}}}, and {{{c=20}}})


{{{t = (-20 +- sqrt( (20)^2-4*-16*20 ))/(2*-16)}}} Plug in a=-16, b=20, and c=20




{{{t = (-20 +- sqrt( 400-4*-16*20 ))/(2*-16)}}} Square 20 to get 400  




{{{t = (-20 +- sqrt( 400+1280 ))/(2*-16)}}} Multiply {{{-4*20*-16}}} to get {{{1280}}}




{{{t = (-20 +- sqrt( 1680 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-20 +- 4*sqrt(105))/(2*-16)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{t = (-20 +- 4*sqrt(105))/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-20 + 4*sqrt(105))/-32}}} or {{{t = (-20 - 4*sqrt(105))/-32}}}



Now break up the fraction



{{{t=-20/-32+4*sqrt(105)/-32}}} or {{{t=-20/-32-4*sqrt(105)/-32}}}



Simplify



{{{t=5 / 8-sqrt(105)/8}}} or {{{t=5 / 8+sqrt(105)/8}}}



So these expressions approximate to


{{{t=-0.65586884574495}}} or {{{t=1.90586884574495}}}




Since a negative time doesn't make sense, our only solution is:

{{{t=1.90586884574495}}}


So you are correct