This problem is a difference-of-two-squares problem, as shown by the squared first term, the missing middle term, and the negative square as the last term.
You solve this by finding the square roots of both terms and use those to find the roots.
[tex]49x^{2}-100=0\\\\
\sqrt{49x^{2}}=(+-)~7x\\
\sqrt{100}=(+-)~10\\\\
(7x+(+10))(7x+(-10))=0\\
(7x+10)(7x-10)=0\\\\
7x+10=0\\
7x=-10\\
x=\frac{-10}{7}\\\\
7x-10=0\\
7x=10\\
x=\frac{10}{7}\\\\
Thus,~x=\frac{-10}{7} ~and~x=\frac{10}{7}[/tex]