The root of the equation 2(1+i)x2−4(2−i)x−5−3i=0,where i=−1, which has greater modulus, is

The given equation is 2(1+i)x2−4(2−i)x−5−3i=0⇒ x=4(2−i)±16(2−i)2+8(1+i)(5+3i)4(1+i)             =4(2−i)±48-64i+40-24+64i4(1+i)             =4(2−i)±644(1+i)=8-4i±84(1+i) =−i1+i or 4−i1+i=−1−i2 or 3−5i2 Now,−1−i2=14+14=12  and 3−5i2=94+254=172Also, 172>12Hence, required root is 3−5i2