Find all integers $x$ such that $x^2+3x+24$ is a perfect square. Find all integers $x$ such that $x^2+3x+24$ is a perfect square.  
My attempt:  
$x^2+3x+24=k^2$
$3(x+8)=(k+x)(k-x)$  
Now, do I find solution treating cases? But that doesn't seem very easy. Please help.

Best way is.... 
$$x^2 + 3x + 24 = k^2 \Rightarrow x^2 + 3x + 24-k^2 = 0\\$$
$$\triangle = b^2 - 4ac = 3^2 - 4(1)(24-k^2) = 9 - 96 + 4k^2 = 4k^2 - 87 = n^2\\$$
$$\to 4k^2 - n^2 = (2k-n)(2k+n) = 87 \\$$
$$\to 2k+n = 29, 2k-n = 3 \to 4k = 32 \to k = 8$$. Can you finish it?
