$\dfrac{3}{35}+\dfrac{3}{63}+\dfrac{3}{99}+...+\dfrac{3}{x(x+2)}=\dfrac{24}{35}$
$↔ \dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{x(x+2)}=\dfrac{24}{35}$
$↔ \dfrac{3}{2}\Bigg[\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{x(x+2)}\Bigg]=\dfrac{24}{35}$
$↔ \dfrac{3}{2}\Bigg(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\Bigg)=\dfrac{24}{35}$
$↔ \dfrac{3}{2}.\Bigg(\dfrac{1}{5}-\dfrac{1}{x+2}\Bigg)=\dfrac{24}{35}$
$↔ \dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}$
$↔ \dfrac{x-3}{5(x+2)}=\dfrac{16}{35}$
$↔ 35(x-3)=16.5(x+2)$
$↔ 35x-105=80x+160$
$↔ 45x=-265$
$↔ x=-\dfrac{53}{9}$
