Question 1119463
Let m=the smallest of the three consecutive integers<br>

Then   {{{ (m)(m+1) + (m+1)(m+2) = 8 }}} <br>

Which reduces to   {{{ m^2 + 2m - 3 = 0 }}}

Solutions are  {{{ m = (-2 +- sqrt(16))/2 }}} 

Potential solutions are  m = -3  and m = 1.
<br>
Check them:
m=-3:    (-3)(-2) + (-2)(-1) = 6+2 = 8    (  {{{  m = -3 }}} is a solution, and the corresponding integers are {{{ highlight(matrix(1,4, "-3, ", "-2, ", "and ", "-1")) }}} )

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m=1:   (1)(2) + (2)(3) = 2 + 6 = 8   (  {{{  m = 1 }}} is a solution, and the corresponding integers are {{{ highlight(matrix(1,4, "1, ", "2, ", "and ", "3")) }}} )

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Tutor ankor@dixie-net made this mistake:
The line
a^2 + a + a^2 + 2a + 1 + 2 = 8  
should be  {{{ a^2 + a + a^2 + 2a + green(a) + 2 = 8 }}}  that is why he did not find the 2nd solution.  


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@MathTherapy - thanks for sharing that most efficient solution!