The root of the equation $2\left( 1+i \right) { x }^{ 2 }-4(2-i)x-5-3i=0 $ which has greater modulus is

Solving 2(1+i)x^2 - 4(2-i)x - (5+3i) = 0 using the quadratic formula x = [-b +/- sqrt(b^2 - 4ac)] / 2a. Discriminant D = [4(2-i)]^2 - 4(2+2i)(-5-3i) = 16(4-1-4i) - 8(1+i)(-5-3i) = 16(3-4i) - 8(-5-3i-5i+3) = 48-64i - 8(-2-8i) = 48-64i + 16 + 64i = 64. Roots are [4(2-i) +/- 8] / [4(1+i)] = (8-4i +/- 8) / (4+4i). Root 1: (16-4i)/(4+4i) = (4-i)/(1+i) = (4-i)(1-i)/2 = (4-4i-i-1)/2 = (3-5i)/2. Root 2: (-4i)/(4+4i) = -i/(1+i) = -i(1-i)/2 = (-i-1)/2. Modulus of (3-5i)/2 is sqrt(9+25)/2 = sqrt(34)/2. Modulus of (-1-i)/2 is sqrt(2)/2. The first root has the greater modulus.