When factoring trinomials will there ALWAYS be 2 numbers that multiply to equal c and add to equal b?

No, not always... this poly does not factor over the integers ( Its called "Prime " over the integers ).

The two roots are around 7.6 and -2.6 and are irrational...

Yes, but they may not be real, rational numbers.

They will be real and rational only if the discriminant is real, perfect square.

Given a trinomial of the form: ax² + bx + c, the discriminant is b²-4ac

If the discriminant < 0 there are no real factors

If the discriminant = 0 there are two equal real factors

if the discriminant > 0 there are two unequal real factors

if the discriminant is > 0 and a perfect square, the real factors are rational

if the discriminant is > 0 and is not a perfect square, the real factors are irrational.

In your case, the discriminant is (-5)² - 4(1)(-20) = 25+80 = 105

The discriminant is > 0 so the factors are real

The discriminant is not a perfect square so the factors are irrational

The factors are the real, irrational numbers (5+√105)/2 and (5-√105)/2

a²-5a-20 = [a - (5+√105)/2][a - (5-√105)/2]

-(5+√105)/2 - (5-√105)/2 = -5

(5+√105)/2 * (5-√105)/2 = (25 - 105)/4 = -20

...

YES! There will always be a way to write it as (x + a)(x+b). The only problem is, a and b may be irrational numbers athat you just can't "see" to start with, or they may even be imaginary. For your problem,

a^2 - 5a - 20 = 0

use the quadratic formula to get x = (5 + √105 ) / 2 and x = (5 - √105) / 2 so you can factor

a^2 - 5a - 20 = (x + (5-√105)/2) * (x + (5 + √105)/2)

So technically it can always be factored (assuming the solution set is NOT empty!), but the factorization may be just too hard to see or too difficult to be of use.

Two numbers must equal AC when multiplied (here that is 1x20)

And they must add to give B (here 5)

So it can be:

1 20

2 10

4 5

None of these work, therefore it will not factorise.

From here you can use the quadratic formula (look it up) or completing the square to give an answer (although it will not be an integer (whole number))

No - in fact, real integer roots for quadratic equations are relatively rare. Your equation may have irrational roots or complex roots. Use the quadratic formula if you are having a hard time finding roots manually.