The given equation is,
27 x 2 −10x+1=0
Comparing the given equation with a x 2 +bx+c=0 , we get
a=27 b=−10 c=1
D= b 2 −4ac = ( −10 ) 2 −4×27×1 =−8
The required solutions are,
x= −b± D 2a = −( −10 )± −8 2×27 = 10±2 2 i 54 = 5 27 ± 2 27 i
Therefore, the solutions to the equation are 5 27 ± 2 27 i .