Answer
$ \displaystyle \frac{9}{2b+9},\quad b\neq\frac{1}{6}$
Work Step by Step
$\displaystyle \frac{22b+15}{12b^{2}+52b-9}+\frac{30b-20}{12b^{2}+52b-9}-\frac{4-2b}{12b^{2}+52b-9}$
... To add/subtract rational expressions with the same denominator,
add/subtract numerators and place the sum/difference over the common denominator.
$= \displaystyle \frac{22b+15+30b-20-(4-2b)}{12b^{2}+52b-9}$
$= \displaystyle \frac{22b+15+30b-20-4+2b}{12b^{2}+52b-9}$
$= \displaystyle \frac{54b-9}{12b^{2}+52b-9}$
Factor what we can,
... $54b-9=9(6b-1)$
... for $ax^{2}+bx+c$ we search for factors of $ac$ whose sum is $b$
... if sucessful, rewrite $bx $ and factor in pairs.
... here, factors of $12\cdot(-9)=2\times 2\times 3\times 3\times 3$
... whose sum is $52$ are $+54=2\times 3\times 3\times 3$ and $-2$
$12b^{2}+52b-9=12b^{2}+54b-2b-9$
$=6b(2b+9)-(2b+9)$
$=(2b+9)(6b-1)$
$= \displaystyle \frac{9(6b-1)}{(2b+9)(6b-1)}$
... common factors cancel,
... exclude b for which $6b-1=0$
=$ \displaystyle \frac{9}{2b+9},\quad b\neq\frac{1}{6}$
