16u$^2$-25v$^2$
6t^2−2t−12
25y$^2$-9x$^2$
lcm of 14.5 and 5.5
16p$^2$-25
prove that cot/1+tanθ=cotθ-1/2-sec²θ
25a$^2$-4
-11+x=-5+x
9n$^2$-1