<s>[INST] Why, my result has the opposite sign to the correct result? I found this exercise of factoring:
$$5a^2 + 3a - 2$$ is a trinomial of the form:
$$ax^2 + bx + c  = \frac{(ax + p)(ax + q)}{a}$$
I made this:
$$3 = p + q$$
$$-10 = pq$$
then, I need to get the values of $p$ and $q$.
$$p = -\frac{10}q$$
replace:
$$3 = -\frac{10}q + q$$
$$q^2 -3q - 10 = 0$$
$$(q - 5)(q + 2)$$ That is two possible values of $q$
I need to check the correct value now:
$$3 = p - 5\implies p = -8$$ but $8 \times 5 \neq -10$, so the correct value of $q$ is $2$.
Replace to check if is true:
$$3 = p - 2\implies p = 1$$ but $1 \times -2 \neq -10$
In the end, the correct values impressively were $5$ and $2$, which I had obtained in the equation of $q$, but with a changed sign.
$$p = 5, q = -2$$
So, what am I doing wrong factoring this expression?
Is it a property that the result was that? or was it coincidence
PD: the total factoring was: $(a + 1)(5a - 2)$
 [/INST] 
the total factoring was: $\;(a + 1)(5a - 2)$

You could get that directly by finding the roots using the quadratic formula $\,\dfrac{-3 \pm \sqrt{49}}{10}=\begin{cases}-1 \\ 2/5\end{cases}\,$.
Then the quadratic factors as $5\big(a-(-1)\big)\left(a - \dfrac{2}{5}\right)=(a+1)(5a-2)\,$.

[ EDIT ] In the general case $\,ax^2 + bx + c  = \dfrac{(ax + p)(ax + q)}{a}\,$ the factorization follows directly from the quadratic formula as well:

$$
\begin{align}
ax^2 + bx + c = a(x-x_1)(x-x_2) &= a\left(x-\frac{-b - \sqrt{\Delta}}{2a}\right)\left(x-\frac{-b + \sqrt{\Delta}}{2a}\right) \\[10px]
 &= \frac{\left(ax + \cfrac{b + \sqrt{\Delta}}{2}\right)\left(ax + \cfrac{b - \sqrt{\Delta}}{2}\right)}{a}
\end{align}
$$
</s>