I need to make "v" the subject of the formula, where v > 0

m = m(subscript 0)/sqrt(1-(v/c)^2)

m = the mass of a particle
m(subscript 0) = the mass of the particle at rest
v = the speed of the particle
c = the speed of light

v > 0

Make "v" the subject of the equation.

The result should look like:

v = (c/m)sqrt(m^2-m(subscript 0)^2)

The problem is... how do you get to that, step by step?

m = m(subscript0) / sqrt[1-(v/c)^2]
m = m(subscript0) / sqrt[1-(v^2/c^2)]
m = m(subscript0) / sqrt[(c^2/c^2)-(v^2-c^2)]
m = m(subscript0) / sqrt[(c^2-v^2)/c^2]
m = m(subscript0) / sqrt(c^2-v^2)/c
m = m(subscript0){c/[sqrt(c^2-v^2)]}
m = cm(subscript0) / sqrt(c^2-v^2)
m[sqrt(c^2-v^2)] = cm(subscript0)
sqrt(c^2-v^2) = cm(subscript0) / m
c^2-v^2 = [cm(subscript0)/m]^2
c^2-v^2 = (c^2)[m(subscript0)^2] / m^2
-v^2 = {c^2[m(subscript0)^2]/m^2}-c^2
-v^2 = {c^2[m(subscript0)^2]/m^2} - [(c^2m^2)/m^2]
-v^2 = {c^2[m(subscript0)^2-c^2m^2} / m^2
-v^2 = (c^2)[m(subscript0)^2-m^2) / m^2
v^2 = (c^2)[-m(subscript0)^2+m^2] / m^2
v^2 = (c^2)[m^2-m(subscript0)^2] / m^2
v = (c/m)sqrt[m^2-m(subscript0)^2]

sqrt[(c^2-v^2)/c^2] means the same thing as sqrt(c^2-v^2) / sqrt(c^2). In the denominator, the sqrt and the exponent cancel out and you're left with just c. √Fraction = √Numerator / √Denominator