Totally lost. ??
I assume you want to simplify this expression?
$\displaystyle \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{4x^2 + 6x -4}{4x-28}$
Start off by factorising your polynomial expressions on the right:
$\displaystyle \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{2(x+2)(2x-1)}{4(x-7)}$
Multiply the two fractions:
$\displaystyle \frac{6(x-7)(x+2)(2x-1)}{4(x+2)(2x-1)(x-7)}$
Cancel out factors in denominator and numerator:
$\displaystyle \frac{6}{4} = \frac{3}{2}$