<s>[INST] How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be multiplied together, (the $(2)(-3)$, rather than $6(-1)$?
When following the factors $-1$ and $6$.  I have 
$(2x^2-1x)(6x-3)$ 
$x(2x-1)+3(2x-1)$
Is this correct, if not; what is the best way to solve a leading coefficient when factoring? 
 [/INST] $$2x^2 + 5x − 3 = 0$$
$$ac = 2(-3) = -6$$
$$\text{$^-1\times 6 =\phantom .^-6 \ $  and  $ \ ^-1+6 = 5$}$$

$-1$ and $6$ are correct. What you did after that is wrong.
Here are two methods that I know of for proceeding from $-1$ and $6$.
Method 1. Replace $5x$ with $-1x+6x$ and factor-by-pairing-off.
\begin{array}{c}
   2x^2 + 5x − 3 \\
   2x^2 -1x + 6x - 3 \\
   (2x^2 -1x) + (6x - 3) \\
   x(2x-1) + 3(2x-1) \\
   (x+3)(2x-1)
\end{array}
Method 2. Write out $(ax-1)(ax+6)$ and then divide out the greatest common divisors.
\begin{array}{c}
   (2x-1)(2x+6) &\{\gcd(2,-1)=1 \ \text{and} \ \gcd(2,6)=2\}\\
   \dfrac{(2x-1)}{1} \cdot \dfrac{(2x+6)}{2} \\
   (2x-1)(x+3)
\end{array}
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