What are x and y if (6-yi)(x+2i)=12-5i?

To determine x and y, we'll have to perform the
multiplication from the left side:

(6-yi)(x+2i) = 6x + 12i
- xyi - 2y`i^(2)`

But `i^(2)` =
-1

(6-yi)(x+2i) = 6x+2y + i(12 -
xy)

Now, we'll equate the real parts from both
sides:

6x+2y = 12 => 3x + y = 6 => y = 6 -
3x

We'll equate the imaginary parts from both
sides:

12 - xy = -5 => xy = 12+5 => xy =
17

x(6 - 3x) = 17

We'll remove
the brackets:

6x - 3`x^(2)` - 17 =
0

3`x^(2)` - 6x + 17 = 0

We'll
apply quadratic formula:

x1 = (6 + `sqrt(36 - 204)`
)/6

x1 = (6 + 2i`sqrt(42)`
)/6

x1 = (3 + i`sqrt(42)`
)/3

x2 = (3 - i`sqrt(42)`
)/3

y1 = 6 - 3x1 => y1 = 6 - 3 -
i`sqrt(42)`

y1 = 3 -
i`sqrt(42)`

y2 = 6 - 3 +
i`sqrt(42)`

y2 = 3 +
i`sqrt(42)`

Therefore, the values of x and y
are: x1 = (3+i`sqrt(42)` )/3 ; y1 = 3 - i`sqrt(42)` ; x2 = (3 - i`sqrt(42)` )/3 ; y2 = 3
+ i`sqrt(42)` .