Question 89292
{{{h = -16t^2 + 70t + 4}}}


{{{0 = -16t^2 + 70t + 4}}} Set h equal to zero


Now 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+70*t+4=0}}} ( notice {{{a=-16}}}, {{{b=70}}}, and {{{c=4}}})


{{{t = (-70 +- sqrt( (70)^2-4*-16*4 ))/(2*-16)}}} Plug in a=-16, b=70, and c=4




{{{t = (-70 +- sqrt( 4900-4*-16*4 ))/(2*-16)}}} Square 70 to get 4900




{{{t = (-70 +- sqrt( 4900+256 ))/(2*-16)}}} Multiply {{{-4*4*-16}}} to get {{{256}}}




{{{t = (-70 +- sqrt( 5156 ))/(2*-16)}}} Combine like terms in the radicand (everything under the square root)




{{{t = (-70 +- 2*sqrt(1289))/(2*-16)}}} Simplify the square root




{{{t = (-70 +- 2*sqrt(1289))/-32}}} Multiply 2 and -16 to get -32


So now the expression breaks down into two parts


{{{t = (-70 + 2*sqrt(1289))/-32}}} or {{{t = (-70 - 2*sqrt(1289))/-32}}}



which split up to



{{{t=-70/-32+2*sqrt(1289)/-32}}} or {{{t=-70/-32-2*sqrt(1289)/-32}}}



and simplify to



{{{t=35 / 16-sqrt(1289)/16}}} or {{{t=35 / 16+sqrt(1289)/16}}}



Which approximate to


{{{t=-0.05641538387703}}} or {{{t=4.43141538387703}}}


Since a negative time doesn't make sense, we must discard the first solution. 


So our solution is {{{t=4.43141538387703}}}


which rounds to {{{t=4.431}}}


So it will take close to 4 and a half seconds to reach the ground