How do I solve the formula #16t^2 - vt - 40 = 0# for #t#?

1 Answer
Steve M
Jun 16, 2018

# t = (v +-sqrt(v^2+2560))/32 #

Explanation:

We have:

# 16t^2-vt-40 = 0 #

We can complete the square, by first dividing by the coefficient of #t^2#

# :. 16{t^2-v/16t-40/16} = 0 #

# :. t^2-v/16t-5/2 = 0 #

Then we factor half the coefficient of #t#:

# :. (t-v/32)^2-(v/32)^2-5/2 = 0 #

Which we can now readily rearrange for #t#:

# :. (t-v/32)^2 = v^2/1024+5/2 #

# :. (t-v/32)^2 = v^2/1024+2560/1024 #

# :. (t-v/32)^2 = (v^2+2560)/1024 #

# :. t-v/32 = +-sqrt((v^2+2560)/1024) #

# :. t-v/32 = +-sqrt(v^2+2560)/32 #

# :. t = v/32 +-sqrt(v^2+2560)/32 #

# :. t = (v +-sqrt(v^2+2560))/32 #

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