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: $(a + 1)(5a - 2)$
 A: The mistake is when you stated that $(q-5)(q+2)=0$ has solutions $q=-5,q=2$. 
Instead, you have $$q-5=0\quad\text{and}\quad q+2=0$$ giving $q=5$ and $q=-2$ and so substituting gives $$3=p+5\implies p=-2\implies pq=-2(5)=-10$$ and $$3=p-2\implies p=5\implies pq=5(-2)=-10$$ as desired.
A: 
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:

$$
\begin{align}
ax^2 + bx + c = a(x-x_1)(x-x_2) &= a\left(x-\frac{-b - \sqrt{\Delta}}{2a}\right)\left(x-\frac{-b + \sqrt{\Delta}}{2a}\right) \\[10px]
 &= \frac{\left(ax + \cfrac{b + \sqrt{\Delta}}{2}\right)\left(ax + \cfrac{b - \sqrt{\Delta}}{2}\right)}{a}
\end{align}
$$
A: Note that $$5a^2 + 3a - 2= (5a+p)(a+q)=5a^2+(p+5q)a+pq$$
Thus $p+5q=3$ and $  pq=-2$
which implies $q=1$ and $p=-2$
Therefore the factorization is
$$5a^2 + 3a - 2=(5a+p)(a+q)=(5a-2)(a+1)$$
A: I'm satisfied with TheSimplyFire's answer for your question. Also look at the equation you set up:
$$ax^2 + bx + c  = \frac{(ax + p)(ax + q)}{a}$$ Now look at the coefficients of $x$: It is $a$ in the both factors of $x$ on RHS. Thus, $p$ and $q$ should be interchangeable and factors belong to only one of two factor pairs of $-10$. That means, $p\times(q)=2\times(-5)$ or $p\times(q)=5\times(-2)$ or $q\times(p)=2\times(-5)$ or $q\times(p)=5\times(-2)$.
Note: Your equation setup can be consider as general form. However, if $a$ in $ax^2 + bx + c$ is a prime number like $5$, you do not have to divide the RHS of the equation again by $a$. Instead, you can setup it as: $$ax^2 + bx + c  = (ax + p)(x + q)$$ This is because any prime number $p$ has one distinctive factor pair: $p\times1$.  
