First, get rid of that pesky 1/2. Fractions are usually difficult to work with, but in this case we can factor out the 1/2 and not have to deal with them at all:
f(x)=1/2(x^2+5x-3)
Now we can ignore the 1/2 and focus on the good stuff:
x^2+5x-3
We must ask ourselves, "Is there a number pair that multiplies to -3 and adds to 5?" We will quickly discover that no such pair exists (no rational numbers meet these criteria), and indeed, we must use the quadratic formula. This is due to fact that the determinant b^2-4ac is not a perfect square.
This equation will have roots at:
x=(-b+-sqrt(b^2-4ac))/(2a)
=(-5+-sqrt(5^2-4(1)(-3)))/(2(1))
=(-5+-sqrt(25+12))/(2)
->x=(-5+-sqrt(37))/(2)
If x=a is a root of a polynomial, then x-a=0, and (x-a) is a factor of the polynomial. So if we want to factor x^2+5x-3, we'll need the roots to look something like x-"blah blah"=0.
Let's get to it. We have:
x=(-5+-sqrt(37))/(2)
Doing some algebra, we can simplify this to:
2x+5=+-sqrt(37)
We have two cases: sqrt(37) and -sqrt(37). We'll work with them both:
2x+5=sqrt(37)->2x+5-sqrt(37)=0
2x+5=-sqrt(37)->2x+5+sqrt(37)=0
So the two factors of our quadratic are 2x+5-sqrt(37)=0 and 2x+5+sqrt(37)=0, and that means we're done.
But note that multiplication of these monomials leads to 4x^2 not 1/2x^2, so we should divide the product by 8. Hence,
f(x)=1/8(2x+5-sqrt(37))(2x+5+sqrt(37))
