solve :(2 + i)x2 - (5 -i)x + 2(1-i) = 0
(2 + i )x2 - ( 5 - i )x + 2 ( 1 -i) = 0
=> 2x2 + ix2 - 5x +ix + 2 - 2i =0
=> 2x2 -5x +2 = -ix2 - ix +2i
=>( x-4 )( x-1) = -i (x2 + x -2)
=>(x-4)(x-1) = -i (x+2)(x-1)
=> (x-4)/ (x+2) = -i
sq both sides we get ;;........
(x-4)2 / (x+2)2 = i2
(x-4)2 = - (x+2)2
x2 + 16 - 8x = -x2 - 4 - 4x
2x2 +20 -4x =0
x2 - 4x +10=0
sovle it furthur by quadratic formula...........
thumps up plzz................