Question

30x^2-47x+14=0

Answer 1

\(\large\color{midnightblue}{30x^2-47x+14=0 }\)
\(\large\color{midnightblue}{(5x-2)(6x-7)=0 }\)

and you got it from here.....

Answer 2

ac method, where AC=420. two num that "x" together to equal 420 but add to = -47 are -12 and -35. so then you get 30x^2-12x-35x+14=0. then factor by grouping and solve for zeros

Answer 3

that is exactly where I kind of got confused, I would set each part equal to zero correct?