Apparent Weight and Circular Motion

A guy reaches the bottom of roller coaster with radius r = 51.0m. How fast does he need to travel for his apparent weight to be three times his actual weight?

Here's what I did:

Fnet = n - w
ma = n - w
ma = 3(-w) - w
(mv^2)/r = -4mg
v^2/r = -4g
v = square root(-4gr)

Obviously you can't square root of a negative but I don't understand what I'm doing wrong

W_a is apparent weight
W_g is weight due to gravity (which is mg)

???what velocity cause W_a=3(W_g) to be true?

W_a= mg+ (F_net_due to the acceleration)

(F_net_due to the acceleration) = ma = (mv^2)/r

W_a = mg + (mv^2)/r

3(W_g)=mg+(mv^2)/r <<< Just substituted for W_a

3(mg)=mg+(mv^2)/r <<< Just sub W_g

3(g)=g+(v^2)/r <<
(3g-g)r=(v^2) << v=sqrt(2gr)

At the bottom:-
=>F(net) = mv^2/r + mg
=>3mg = mv^2/r + mg
=>mv^2/r = 2mg
=>v = √2gr
=>v = √[2 x 9.8 x 51]
=>v = 31.62 m/s