My question is 2x^2+13x+6=0 there are no numbers that multiply to 6 and can add to 13. i'm so !

Look

2x^2+13x+6
factored into
(ax+b)(cx+d)
we need to find a,b,c,d such that when multipied we get 2x^2+13x+6
first
we know that the first number hast to be 2
(2x+somethin)(1x+sometin)
and the end numbers have to multiply to 6
if we try 1 and 6
(2x+1)(x+6)=
2x^2+x+12x+6=
2x^2+13x+6
correct

so
2x^2+13x+6=0=(2x+1)(x+6)

therefor assume
2x+1=0 and x+6=0

2x+1=0
minus 1
2x=-1
divide 2
x=-1/2

other one
x+6=0
mnus 6
x=-6

x=-6 or -1/2

It is the Prophase as the first stage of four.