For factoring simple quadratic equations, it's simply a matter of remembering simple forms and tricks. For example, one of the easiest quadratic factorings is the difference of squares
$a^2 -b^2 = (a+b)(a-b)$
Here are several examples of difference of squares factoring:
$x^2 - 1 = (x+1)(x-1) \\ 9x^2 -4 = (3x+2)(3x-2) \\ 36x^2 - 25y^2 = (6x +5y)(6x-5y)$
The next simplest factoring is quadratics without constant terms, which boils down to just spotting common factors. Ex: $9x^2 +3x = 3x(3x+1)$
Next there is factoring quadratics with a leading coefficient of 1, which amounts to pairing factors:
If $x^2 + hx +k = (x+a)(x+b)$, then $k = ab$ and $h = a+b$. This comes from a simple expansion of $(x+a)(x+b) = x^2 + ax + bx + ab = x^2 + (a+b)x + ab$
For example, $x^2 + 20x + 36$. The factors of $36$ are $(1,36) \ (2,18) \ (3,12) \ (4,9) \ (6,6)$ Notice that $20 = 2+18$. Thus, $x^2 + 20x + 26 = (x+2)(x+18)$. This can always be checked by re-expanding.
This even works with negative terms. For example, $x^2 - 4x -12$ The factors of $-12$ are $(-1, 12) \ (-2,6) \ (-3, 4) \ (-4,3) \ (-6, 2) \ (-12, 1)$. Notice that $-6 + 2 = -4$. Thus, $x^2 - 4x -12 = (x-6)(x+2)$
The last case is when the leading coefficient is anything other than $1$. This can be done with a method similar to the one above, but with slightly more computation. For any quadratic polynomial $ax^2 + bx + c$, if it has an expansion in integers $(px + q)(rx +s)$, then $a = pr$, $c = qs$, and $b = (ps + qr)$. Thus, this method of factoring involves finding all the factors of $a$ and $c$, and pairing them in such a way as to equal the middle term $b$. It's easier to demonstrate than to explain.
Let's use your example: $4x^2 - 4x - 3$. The factors of $4$ are $(1,4) \ (2,2)$ and the factors of $-3$ are $(-1,3) \ (1, -3)$. Let's begin pairing exhaustively.
$(1)(-1) + (4)(3) = 11 \\
(1)(3) + (4)(-1) = -1 \\
(1)(1) + (4)(-3) = -11 \\
(1)(-3) + (4)(1) = 1 \\
(2)(-1) + (2)(3) = 4 \\
(2)(1) + (2)(-3) = -4$
Every one of these expressions is equivalent to (a factor of $a$)(a factor of $b$) + (the corresponding factor of $a$)(the corresponding factor of $b$)
Notice that the last expression is the one we need. We need to take the numbers $2, 1, 2, -3$ and plug them into $(\_x +\_)(\_x+\_)$ such that the product is equal to $4x^2 -4x -3$
We know the coefficients of the $x$ terms will be $(2x+\_)(2x+\_)$, because they must multiply to be $4$. From here, we would place the $1, -3$ in the expression such that they match up with the appropriate term to multiply by. However in this case since it's just two $2$'s, the answer is $4x^2 - 4x -3 = (2x+1)(2x-3)$. Check this by expanding back out.
Another example of this method: $3x^2 -7x + 2$. The factors of $3$ are $(1,3)$ and the factors of $2$ are $(1,2)$. Notice that the middle term is negative, though, so we must use $(-1, -2)$ as our factors of $2$. Now we exhaustively pair.
$(1)(-2) + (3)(-1) = -5 \\
(1)(-1) + (3)(-2) = -7$.
The last expression is the one we need. Thus, the factored expansion is $(x+\_)(3x+\_)$. Notice here that for the expansion to be correct, the $3$ must multiply by $-2$. Thus, we plug in into the factor $(x+\_)$, otherwise $-2$ and $3$ would not be multiplied.
Thus, $3x^2 - 7x + 2 = (x-2)(3x-1)$. Check this by expanding back out.
If the method of factor-pairing by exhaustion reveals that there is no possible combination of factors which will sum to the middle term, then the expression has no factorization in integers.
Best Answer
$2x^2-x+3=2(x^2-\frac{1}{2}x+\frac{3}{2})$ where the discriminant is $\Delta=(\frac{-1}{2})^2-4\times 1\times \frac{3}{2}=\frac{1}{4}-6<0$. Therefore $x^2-\frac{1}{2}x+\frac{3}{2}$ has no real roots and so it is an irreducible polynomial over $\mathbb{R}$.