Find the primes $p$ such that the equation: $x^{2} + 6x + 15 = 0$ has a solution modulo $p$ I need to solve this question: 
Find the primes $p$ such that the equation: $x^{2} + 6x + 15 = 0  $ has a solution modulo $ p $.
My approach was: I checked for $p = 2$ and there is no solution.
Now if $p  \neq 2 $ so the equation has a solution $\iff  6^{2} -4 \times 15 = -24$ is a square modulo $p$.
Now $-24 $ is a square modulo $p \iff \left(\frac{-24}{p}\right) = 1$ (Legendre symbol) $ \iff \left(\frac{-1}{p}\right)  \left(\frac{2}{p}\right)  \left(\frac{3}{p}\right) = 1.$
And now there are many  cases to check and I'm not sure how to do it...
For example - case 1:
$ \left(\frac{-1}{p}\right) =1 $ and $  \left(\frac{2}{p}\right) =1 $ and $ \left(\frac{3}{p}\right) = 1$ : 
$ \left(\frac{-1}{p}\right) =1  \iff p \equiv 1 (4) $  ,
$  \left(\frac{2}{p}\right) =1  \iff p \equiv 1, -1 (8) $ , 
$  \left(\frac{3}{p}\right) = 1 \iff p \equiv 1, -1 (12) $ 
so how do I combine these results for case 1? and after that, do I really need now to check all the other cases - that 2 of the Legendre symbols are $(-1)$ and one is $1$ ? 

$x^2+6x+15 \equiv 0\pmod{p}\Rightarrow(x+3)^2+6\equiv0\pmod{p}\Rightarrow (x+3)^2\equiv-6\pmod{p}$.
In other words, $y^2\equiv -6\pmod{p}$ or to rephrase the argument, $\left(\frac{-6}{p}\right)=1$
