<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] We need to find two numbers with product $2(-3) = -6$ and sum $5$.  You are correct that the numbers are $-1$ and $6$.  Splitting the linear term yields
\begin{align*}
2x^2 + 5x - 3 & = 2x^2 - x + 6x - 3 && \text{split the linear term}\\
              & = x(2x - 1) + 3(2x - 1) && \text{factor by grouping}\\
              & = (x + 3)(2x - 1) && \text{extract the common factor}
\end{align*}
which you can verify by multiplying the factors.
You should not write $(2x^2 - 1x)(6x - 3)$ since 
\begin{align*}
(2x^2 - 1x)(6x + 3) & = 2x^2(6x + 3) - 1x(6x + 3)\\
                    & = 12x^3 + 6x^2 - 6x^2 -3x\\
                    & = 12x^3 - 3x\\
                    & \neq 2x^2 + 5x - 3
\end{align*}
Instead, you can write $(2x^2 - 1x) + (6x - 3)$ or $2x^2 - 1x + 6x - 3$.  
Also, you should be including equals signs since you are asserting that 
$$2x^2 + 5x - 3 = 2x^2 - x + 6x - 3$$
</s>