(16-2i)/(5+i)=3-i

Solution for (16-2i)/(5+i)=3-i equation:

(16-2i)/(5+i)=3-i

We move all terms to the left:

(16-2i)/(5+i)-(3-i)=0


Domain of the equation: (5+i)!=0

We move all terms containing i to the left, all other terms to the right

i!=-5

i∈R


We add all the numbers together, and all the variables

(-2i+16)/(i+5)-(-1i+3)=0

We get rid of parentheses

(-2i+16)/(i+5)+1i-3=0

We multiply all the terms by the denominator


(-2i+16)+1i*(i+5)-3*(i+5)=0

We multiply parentheses

i^2+(-2i+16)+5i-3i-15=0

We get rid of parentheses

i^2-2i+5i-3i+16-15=0

We add all the numbers together, and all the variables

i^2+1=0

a = 1; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·1·1
Δ = -4
Delta is less than zero, so there is no solution for the equation