(1+i)(3-2i)=

Solution for (1+i)(3-2i)= equation:

Simplifying
(1 + i)(3 + -2i) = 0

Multiply (1 + i) * (3 + -2i)
(1(3 + -2i) + i(3 + -2i)) = 0
((3 * 1 + -2i * 1) + i(3 + -2i)) = 0
((3 + -2i) + i(3 + -2i)) = 0
(3 + -2i + (3 * i + -2i * i)) = 0
(3 + -2i + (3i + -2i2)) = 0

Combine like terms: -2i + 3i = 1i
(3 + 1i + -2i2) = 0

Solving
3 + 1i + -2i2 = 0

Solving for variable 'i'.

Factor a trinomial.
(3 + -2i)(1 + i) = 0

Subproblem 1
Set the factor '(3 + -2i)' equal to zero and attempt to solve:

Simplifying
3 + -2i = 0

Solving
3 + -2i = 0

Move all terms containing i to the left, all other terms to the right.

Add '-3' to each side of the equation.
3 + -3 + -2i = 0 + -3

Combine like terms: 3 + -3 = 0
0 + -2i = 0 + -3
-2i = 0 + -3

Combine like terms: 0 + -3 = -3
-2i = -3

Divide each side by '-2'.
i = 1.5

Simplifying
i = 1.5
Subproblem 2
Set the factor '(1 + i)' equal to zero and attempt to solve:

Simplifying
1 + i = 0

Solving
1 + i = 0

Move all terms containing i to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + -1 + i = 0 + -1

Combine like terms: 1 + -1 = 0
0 + i = 0 + -1
i = 0 + -1

Combine like terms: 0 + -1 = -1
i = -1

Simplifying
i = -1
Solution
i = {1.5, -1}