This exercise is not like being told to find the values of x that satisfy something like 7x^2 - 10x - 3 = 0 (in other words, having to find x when y is 0). In this exercise, y is not zero.judocallin said:Am I going to use the form ax^2 +bx+c=0, to solved this? No. There is nothing "to solve", in this manner.
judocallin said:divide 5 by 2 to get "A"
to get "B" subtract -2 by "A"
and -4 is the last term
Good grief. :roll:
It works, in this case, but this trick does not work, in most cases.
giving is the naswer (5/2)x^2-(9/2)x-4 This is the same result that you earlier posted was wrong, according to your professor!
We already told you that it's correct, for the three points that you provided.