Special relativity question physics

Formula:
T=D/V
T inside = t outside * square root [1-(v^2 / c^2)
C = 2.98 * 10^8
Find the velocity that would produce a time dilation factor of .99. This would only produce a time difference of only 1%. To put this in perspective the space shuttles top speed in orbit is 1.4*10^4 m/s.

Please help! I am stuck on this for an hour and I couldn't figure it out

its a Lorentz transformation problem...

dilated time= normal time x gamma

gamma= 1/sqrt(1- (v/c)^2) (one over the square root of v/c squared)
c= the speed of light, 3(10^8) m/s
v= velocity

thats hard to type, but basically you want dilated time to be .99 and normal time to be 1, so you end up with

1= .99/sqrt(1-(v/c)^2)

solve for v

.99= sqrt(1-(v/c)^2)
.9801= 1-(v/c)^2
.0199= (v/c)^2
.1411= v/c
v = .14c = 4.23(10^7) m/s

should be right

substitute c=2.98 * 10^8 into formula, and t=0.99 T=1
1= 0.99/(1-(v/(2.98 * 10^8))^2)^(1/2)
(1-(v/(2.98 * 10^8))^2)^(1/2)=0.99
square both sides
1-v/(2.98 * 10^8))^2=0.9801
v^2/(2.98 * 10^8)^2=0.0199
v^2=0.0199*(2.98 * 10^8)^2
v=(0.0199*(2.98 * 10^8)^2)^(1/2)=4.20*10^7 m/s
=3002.7 times as fast as space shuttles