18c^3 - 21c^2 + 30c - 35
I would solve by grouping:
(18c^3 - 21c^2) + (30c - 35)
Now pull outside the parentheses anything incomming within the parentheses:
18^3 - 21c^2 have 3c^2 in common so pull that out.
30c - 35 have 5 in common so pull that out
Your new equation should look like this:
3c^2(6c - 7) + 5(6c - 7)
Now put the outside parts into one parenthese and the inside parts into one parentheses:
(3c^2 + 5)(6c - 7)
Now set both parts equal to zero and solve for c:
3c^2 + 5 = 0
3c^2 = -5 Subtracted 5 from both sides
c^2 = -5/3 Divided both sides by 5
c = + / - sqrt(-5/3) Square rooted both sides.
Now if your teacher wants you to worry about imaginary numbers you could have c values of -5/3i or 5/3i. If not then say that these dont exist.
Now for the second set of parentheses.
6c - 7 = 0
6c = 7 Add 7 to both sides.
c = 7/6 Divide by 6.
This one is a real number answer.