The equation is tan(30+θ) = 2tan(60-θ), how can it be written in the form of tan^2θ + (6√ 3)tanθ – 5 = 0?

See the derivation in the Explanation Section below.
Suppose that, ##(30+theta)=alpha, &, (60-theta)=beta##, so that,
##alpha+beta=90, or, beta=90-alpha##.
Sub.ing these in the given eqn., we have,
##tanalpha=2tanbeta=2tan(90-alpha)=2cotalpha=2/tanalpha##, or,
##tan^2alpha=2##
##:. tan^2(30+theta)=2##
##:. ((tan30+tantheta)/(1-tan30*tantheta))^2=2##.
##:. ((1/sqrt3+t)/(1-1/sqrt3*t))^2=2, “where, “t=tantheta##
##:. (1+sqrt3*t)^2=2(sqrt3-t)^2##.
##:. 1+2sqrt3*t+3t^2=6-4sqrt3*t+2t^2##.
##:. t^2+6sqrt3*t-5=0##, i.e.,
##tan^2theta+6sqrt3tantheta-5=0##, as desired!
Enjoy Maths.!.