We always have
$(x+a)(x+b) = x(x+b)+a(x+b) = (x\times x)+(x\times b)+(a\times x)+(a\times b) = (x^2)+(b\times x)+(a\times x)+(a\times b)$
$(x^2)+(b\times x)+(a\times x)+(a\times b) = (x^2)+(a\times x)+(b\times x)+(a\times b) = (x^2)+((a+b)\times x)+(a\times b)$
$(x+a)\times (x+b) = (x^2)+((a+b)\times x)+(a\times b)$
Converting your equation to that form gives
$x^2+x-6 = (x^2)+(1\times x)+(-6)$
where we want
$(x^2)+(1\times x)+(-6) = (x^2)+((a+b)\times x)+(a\times b)$
So, we're looking for numbers whose sum is $1$ and whose product is $-6$.
$a+b = 1$ and $ab = -6$
That should be enough for you to figure it out, but in case it's not,
if you just want the answer you can put
"n vf rdhny gb artngvir gjb naq o vf rdhny gb cbfvgvir guerr"
into rot13.com