Question 90300
Are you sure the sum is not 110? If the sum is 116, then you get a decimal answer, which is not an integer.


{{{x^2+(x+1)^2+(x+2)^2=110}}} Start with the given expression


{{{x^2+x^2+2x+1+x^2+4x+4=110}}} Foil


{{{x^2+x^2+2x+1+x^2+4x+4-110=0}}} Subtract 110 from both sides


{{{3x^2+6x-105=0}}} Combine like terms





Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{3*x^2+6*x-105=0}}} ( notice {{{a=3}}}, {{{b=6}}}, and {{{c=-105}}})


{{{x = (-6 +- sqrt( (6)^2-4*3*-105 ))/(2*3)}}} Plug in a=3, b=6, and c=-105




{{{x = (-6 +- sqrt( 36-4*3*-105 ))/(2*3)}}} Square 6 to get 36  




{{{x = (-6 +- sqrt( 36+1260 ))/(2*3)}}} Multiply {{{-4*-105*3}}} to get {{{1260}}}




{{{x = (-6 +- sqrt( 1296 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-6 +- 36)/(2*3)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-6 +- 36)/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (-6 + 36)/6}}} or {{{x = (-6 - 36)/6}}}


Lets look at the first part:


{{{x=(-6 + 36)/6}}}


{{{x=30/6}}} Add the terms in the numerator

{{{x=5}}} Divide


So one answer is

{{{x=5}}}




Now lets look at the second part:


{{{x=(-6 - 36)/6}}}


{{{x=-42/6}}} Subtract the terms in the numerator

{{{x=-7}}} Divide


So another answer is

{{{x=-7}}}


So our solutions are:

{{{x=5}}} or {{{x=-7}}}



Plug in x=5


1st #: {{{x=5}}}


2nd #: {{{y=x+1=5+1=6}}}


3rd #: {{{z=x+2=5+2=7}}}



Plug in x=-7


1st #: {{{x=-7}}}


2nd #: {{{y=x+1=-7+1=-6}}}


3rd #: {{{z=x+2=-7+2=-5}}}



So that means our 3 numbers are 


5,6,7


or 

-7,-6,-5