SOLUTION: Find two consecutive integers such that the square of the sum of the two integers is 11 more than the lesser integer. I'm having a terrible time solving this. According to the b

Question 297402: Find two consecutive integers such that the square of the sum of the two integers is 11 more than the lesser integer.
I'm having a terrible time solving this. According to the book, the answer should be (-2,-1).I believe the formula is:
x = lesser #
x+1 = greater #
(x+x+1)^2=x+11
(2x+1)^2=x+11
4x^2+4x+2=x+11 (subtract the x+11 to bring it to the other side)
4x^2+3x-9=0 (this is also where I get stuck)
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
(x+x+1)^2=x+11
(2x+1)^2=x+11
4x^2+4x+[1]=x+11 (subtract the x+11 to bring it to the other side)
[You missed the 1*1=1 term.]
4x^2+3x-[10]=0
(4x-5)(x+2)=0
4x-5=x=5/4 not an integer.
x+2=0
x=-2 ans.
Proof:
(-2-1)^2=-2+11
-3^2=9
9=9
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