So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

Next, square both sides:

Next, subtract x and add 2 to both sides of the equation:

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

Now you can rewrite the equation as 
Now, apply the Zero Product Property and solve for x as such:

Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>
Answer: y= -3/2x + 2
y= -3/2 + b
5 = -3/2 * -2 + b
5 = 3 + b
2 = b
Switch y and x to find the inverse:
x = 3y - 8
Now just solve for y
y = x/3 + 8/3