The equation is tan(30+) = 2tan(60-) how can it be written in the form of tan^2 + (6 3)tan – 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.!.