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

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

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

Multiply (2 + i) * (3 + -2i)
(2(3 + -2i) + i(3 + -2i)) = 0
((3 * 2 + -2i * 2) + i(3 + -2i)) = 0
((6 + -4i) + i(3 + -2i)) = 0
(6 + -4i + (3 * i + -2i * i)) = 0
(6 + -4i + (3i + -2i2)) = 0

Combine like terms: -4i + 3i = -1i
(6 + -1i + -2i2) = 0

Solving
6 + -1i + -2i2 = 0

Solving for variable 'i'.

Factor a trinomial.
(3 + -2i)(2 + 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 '(2 + i)' equal to zero and attempt to solve:

Simplifying
2 + i = 0

Solving
2 + i = 0

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

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

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

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

Simplifying
i = -2
Solution
i = {1.5, -2}