The expression given is,
[tex]\frac{14x^2-45x-14}{21x^2-50x-16}[/tex]
Simplify
[tex]\begin{gathered} =\frac{\left(7x+2\right)\left(2x-7\right)}{21x^2-50x-16} \\ =\frac{\left(7x+2\right)\left(2x-7\right)}{\left(7x+2\right)\left(3x-8\right)} \\ \mathrm{Cancel\:the\:common\:factor:}\:7x+2 \\ =\frac{2x-7}{3x-8} \end{gathered}[/tex]
Hence, Rational expression in lowest terms is,
[tex]\frac{2x-7}{3x-8}[/tex]
Variable restrictions for the original expression:
[tex]\begin{gathered} \left(7x+2\right)\left(3x-8\right)=0 \\ 7x+2=0,3x-8=0 \\ 7x=-2,3x=8 \\ x=-\frac{2}{7},x=\frac{8}{3} \\ \therefore x\ne\frac{2}{7},\frac{8}{3} \end{gathered}[/tex]
Hence, the answer is
[tex]\begin{equation*} x\ne\frac{2}{7},\frac{8}{3} \end{equation*}[/tex]