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

Explanation

`2(1+ i)x^(2) - 4(2-i)x - 5-3i =0`
`rArr x =(4(2-i) pm sqrt(16(2-i)^(2)+4.2(1+i).(5+3i)))/(4(1+i))`
`= (4(2-i) pm sqrt( 16(4-4i-1)+8(5+3i+5i+3)))/(4(1+i))`
`=(4(2-i)pmsqrt(48-64i+16+64i))/(4(1+i))`
`=(4(2-i)pm8)/(4(1+i))`
`((2-i) pm 2)/(1+i)`
`= (2-i+ 2)/(i),(2-i-2)/(1+i)`
`((4-i)(1-i))/(2),(-i(1-i))/(2)`
`=(3-5i)/(2),-((1+i)/(2))`