Question 1047994
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On a 3 positive consecutive integer the middle number is p. If 3 times the square of the largest number is greater 
than the sum of the other 2 number by 67 find the value of p
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<pre>
So, the numbers are p-1, p, and p+1.
The condition says

{{{3*(p+1)^2}}} - ((p-1) + p) = 67.

Simplify

{{{3p^2 + 6p + 3 - p + 1 - p}}} = 67,

{{{3p^2 + 4p + 4}}} = 67,

{{{3p^2 + 4p - 63}}} = 0.

There is no solution in integer numbers.
</pre>

<U>Answer</U>.  There is no solution in integer numbers.