Biomechanics Quadratic Equation problem

[dsm63]

Homework Statement

1.
Neglecting the height of release, a ball is thrown vertically upwards at 20 m/s, find:
d) time (s) that the projectile is at height 10.5 m.

V1 = 20m/s
a = -9.81m/s^2
d = 10.5m

Homework Equations

d = v1*t + 1/2 a(t^2)

The Attempt at a Solution

subsitutiting i can get the formula to:

10.5 = 20t + 1/2(-9.81)t^2
10.5 -20t +4.905t^2 = 0
4.905t^2 - 20t = -10.5
t^2 - 20/4.905 t = -10.5/4.905

However I am rusty on my quadratic equation work and do not know how to solve from here and cannot seem to figure it out? any help would be awesome Thanks in advance.

 

[tiny-tim]

welcome to pf!

hi dsm63! welcome to pf!

(try using the X² button just above the Reply box )

10.5 -20t +4.905t² = 0

either use the formula [-b ± √(b2 - 4ac)]/2a,

or complete the square

 

[dsm63]

would i be right to assume that
for

[-b ± √(b2 - 4ac)]/2a

a = 4.905
b = -20
c = 10.5

when using this i get

t = 20 ±√19.7747

t = 24.44 or 15.55

this answer does not make sense? i think i have some something wrong..

using a calculator online i recieved

t = .619 or 3.45s

 

[tiny-tim]

(just got up :zzz:)

would i be right to assume that
for

[-b ± √(b² - 4ac)]/2a

that's correct

but you've done -b ± √[(b² - 4ac)/2a] !