2x+3=17
subtract 3 from each side
2x+3 -3 = 17 -3
2x = 14
divide by 2 on each side
2x/2 = 14/2
x = 7
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>
Step-by-step explanation:
All has been answered... send me any questions at comments