The firs thing you need to do in this type of problem is factor it. First, I look at the leading coefficient of x^2. As there is no coefficient given, you can leave it as x:
(x )(x )
I then look at the last number (in this case a 4) The two factors of 4 are +/ 2 and 4 and 1.
(x2)(x+2) or (x4)(x+1) or (x1)(x+4)
If I were to FOIL these though, I would find that I could not get a positive 6x. At this point I would use the quadratic formula:
x=  b +/ (√(b24ac)/2a
x = (6 + √(6^{2}4(1)(4))/2(1) and x = 6  √(624(1)(4))/2(1)
x = 6 + √(36(16))/2 and x = 6  √(52)/2
x = 6 + √(52)/2 and x = 6  √(52)/2
√(52) > √13 * √4 > 2√13
x = (6 + 2√13)/2 and x = (6  2√13)/2
x = 3 + √13 and x = 3  √13 (Exact Answers)
x = .61 and x = 6.6 (Decimal Answers)
Oct 28

Tymon E.