Answer
The solution is $b=2$
Work Step by Step
$\dfrac{5}{b+5}-\dfrac{4}{b^{2}+2b}=\dfrac{6}{b^{2}+7b+10}$
Factor all rational expressions completely, if possible:
$\dfrac{5}{b+5}-\dfrac{4}{b(b+2)}=\dfrac{6}{(b+5)(b+2)}$
Multiply the whole equation by $b(b+5)(b+2)$:
$b(b+5)(b+2)\Big[\dfrac{5}{b+5}-\dfrac{4}{b(b+2)}=\dfrac{6}{(b+5)(b+2)}\Big]$
$5b(b+2)-4(b+5)=6b$
$5b^{2}+10b-4b-20=6b$
Take $6b$ to the left side and simplify:
$5b^{2}+10b-4b-20-6b=0$
$5b^{2}-20=0$
Take $20$ to the right side:
$5b^{2}=20$
Take $5$ to divide the right side:
$b^{2}=\dfrac{20}{5}$
$b^{2}=4$
Take the square root of both sides:
$\sqrt{b^{2}}=\pm\sqrt{4}$
$b=\pm2$
The original equation is undefined for $b=-2$, so the solution is only $b=2$
