Given 5tan2a = 2tana + 6, find a to the nearest tenth of a degree when 180º≤a≤270º

We know that
if  180º≤A≤270º
then 
A belongs to the III quadrant and tan A is positive

we have
5tan²A = 2tanA + 6-------> 5tan²A - 2tanA - 6=0

let
x=tan A
5x²-2x-6=0

using a graph tool----> to resolve the second order equation
see the attached figure
x=-0.91
x=1.314
remember that tan A is positive
the solution is
x=1.314
tan A=1.314
A=arc tan (1.314)--------> A=52.73°-----> A=52.7°
 180º≤A≤270º
then
 A=180+52.7-----> A=232.7°

the answer is
A=232.7°

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