Ummm i am really not sure where u got that answer from sweetheart. what you do is combine like terms. Like all the ones with x^2 go together and all the ones with just x go together. once you do that you get 2x^2+4x-2 because 4x^2 + -2x^ would be 2x^2 and -3x + 7x would be 4x. then you can take out a 2 from the whole problem if you like and that would give you a 2(x^2+2x-1). that would be my final answer.

3(m+3)+4<15=3m hmm... Is even achievable to have > and = in a difficulty? it ought to represent, besides the undeniable fact that [3(m+3)+4)] = 3m, and the respond would desire to be decrease than 15 —–additionally, the respond may well be no answer, you by no skill comprehend. so perhaps that merely skill that the left area can not be greater than 15. So, in case you have been to easily start up putting in numbers, then; (a million) – [3(a million+3)+4)] < 15 = 3m ~ 13 = 3 X (2) - [3(2+3)+4)] < 15 = 3m ~ 19 = 9 X (-a million) - [3(-a million+3)+4)] < 15 = 3m ~ 10 = -3 X (-2) - [3(-2+3)+4)] < 15 = 3m ~ 7 = -6 X (-3) ~ [3(-3+3)+4)] < 15 = 3m ~ 4 = -9 X (a million/3) - [3(a million/3+3)+4)] < 15 = 3m ~ 8 = a million X (a million a million/3) - [3(a million a million/3 +3)+4)] < 15 = 3m ~ 11 = 4 X (2 a million/3) - [3(2 a million/3+3)+4)] < 15 = 3m ~ 20 = 7 X i think no solutions got here up... yet perhaps now you have a greater proper theory on a thank you to do it? wish I helped slightly.