Physics question: particles

Particles of charge Q and 3Q are placed on the x-axis at x=-L and x=+L, respectively. A third particle of charge q is placed on the x-axis, and it is found that the total electric force on this particle is zero. Where is the particle?

Answer: x = -0.27L

I would like to know the steps.

kQq/d^2 = k3Qq/(2L-d)^2

(2L-d)^2 = 3d^2

2L-d = +- root3 d = +- 1.732d

2L = 2.732 d or 2L = -0.732d

d = 0.732 L or d = -2.73 L from the charge Q

so d = - 0.27 (from the origin)

force between charges = q1*q2*k/r^2
the charge q is at, (-L+x) from Q, and is at (L+x) unit from 3Q,
so at equilibrium, -Q*q*k/(-L+x)^2=q*3Q*k/(L+x)^2; note negative sign on LHS because force is directional.
simplified to -1/(-L+x)^2=3/(L+x)^2
-(L+x)/(-L+x)=sqrt 3
-L-x=-sqrt3 *L +sqrt3 *x
(-1-sqrt3)x=(1-sqrt3)L
x=(1-sqrt3)/(-1-sqrt 3) *L
x=-0.73/2.73 *L
x= -0.27L