Question on complex equations

How would I go about solving this problem?

Find the value of c and d that make the equation true.

2ci + 1 = -d + 6 - ci

i = √-1 (imaginary)

the imaginary numbers (number of "i"s) on both sides of the equation must be equal. That is:

2ci = -ci
2c = -c
3c = 0
c = 0

Plugging in c=0 we get:

1 = -d +6
-5 = -d
d = 5

Therefore c=0, d=5.

d+3ci=5

It's apparent that c must equal i to get rid of the complex number.

so d+3i^2=5
d-3=5
d=8

Put in the imaginary number and square both side of the equation.