Asked May 28, 2011, 09:36 AM — 1 Answer
Find integral values of x such that value of x^2+19x+92 is a perfect square of an integer

 jcaron2 Posts: 983, Reputation: 1034 Senior Member #2 May 28, 2011, 08:53 PM
We can just say

$x^2+19x+92 = n^2$,

Where n is an integer.

Now just use the quadratic equation to solve for x:

$x=\frac{-19 \pm \sqrt{19^2-4(92-n^2)}}{2}$

Plug in whatever values of n you want, and you'll have your possible values of x.

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