The given equation is,
x 2 −2x+ 3 2 =0
2 x 2 −4x+3=0
Comparing the given equation with a x 2 +bx+c=0 , we get
a=2 b=−4 c=3
D= b 2 −4ac = ( −4 ) 2 −4×2×3 =−8
The required solutions are,
x= −b± D 2a = −( −4 )±2 2 i 4 = 2± 2 i 2 =1± 2 2 i
Therefore, the solutions to the equation are 1± 2 2 i .