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°