How do you solve #v^2/25-1=11#?

Question

1287 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-30T12:06:46

#v# is approximately #+-17.32# or #+-sqrt(300)#

To solve for #v# perform the following operations to keep the equation balanced while isolating #v#:

#v^2/25 - 1 + 1 = 11 + 1#

#v^2/25 = 12#

#(v^2/25)25 = 12 * 25#

#V^2 = 300#

#sqrt(V^2) = +-sqrt(300)#

#v = +-17.32#