Question 751433
Solve {{{-16t^2+48t+32=0}}} for t


Use the quadratic formula to solve for t


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


{{{t = (-(48)+-sqrt((48)^2-4(-16)(32)))/(2(-16))}}} Plug in {{{a = -16}}}, {{{b = 48}}}, {{{c = 32}}}


{{{t = (-48+-sqrt(2304-(-2048)))/(-32)}}}


{{{t = (-48+-sqrt(2304+2048))/(-32)}}}


{{{t = (-48+-sqrt(4352))/(-32)}}}


{{{t = (-48+sqrt(4352))/(-32)}}} or {{{t = (-48-sqrt(4352))/(-32)}}}


{{{t = (-48+16*sqrt(17))/(-32)}}} or {{{t = (-48-16*sqrt(17))/(-32)}}}


{{{t = (3-sqrt(17))/2}}} or {{{t = (3+sqrt(17))/2}}}


{{{t = -0.561553}}} or {{{t = 3.561553}}}


Ignore the negative solution (since a negative time doesn't make sense)


So the only solution is approximately {{{t = 3.561553}}}


This means it will take approximately <font color="red">3.561553 seconds</font> for it to hit the ground.