Question 1061592

find three consecutive odd positive integers such that 3 times the sum of all 3 is 26 less than product of the first and second integers
<pre>Let smallest be S
Then others are: S + 2, and S + 4
We then get: 3(S + S + 2 + S + 4) = S(S + 2) - 26
{{{3(3S + 6) = S^2 + 2S - 26}}}
{{{9S + 18 = S^2 + 2S - 26}}}
{{{S^2 + 2S - 26 - 9S - 18 = 0}}}
{{{S^2 - 7S - 44 = 0}}}
(S - 11)(S + 4) = 0
S, or {{{highlight_green(matrix(1,4, Smallest, integer, "is:", 11))}}}       OR         S = - 4 (ignore) 
      {{{highlight_green(matrix(2,3, Middle, "integer:", 13, Largest, "integer:", 15))}}}