<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 have that $(2x+a)(x+b)=2x^2+5x-3$, where $x=-\frac a2$ and $x=-b$ are the solutions you require.
Then $$(2x^2+(a+2b)x+ab=2x^2+5x-3$$ 
 leads to $$a+2b=5; ab=-3$$
We set $b=\frac12(5-a)\to\frac12a(5-a)=-3 \to a^2-5a-6=0$
$$\to a=-1, 6$$
$$\to b= 3, -\frac 12$$
So $$2x^2+5x-3=(2x-1)(x+3)=(2x+6)(x-\frac12)$$
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