Assuming that the equation is:
[tex] \frac{5}{x {}^{2} } + 7x + 12[/tex]
Then the restriction is x cannot be 0, as the denominator of the first term can't be 0.
However, it is more likely that the equation is
[tex] \frac{5}{ {x}^{2} + 7x + 12} [/tex]
And so you'd need to factorise it so it becomes
[tex] \frac{5}{(x + 4)(x + 3)} [/tex]
Similar to above, the denominator can't be 0, hence in this case the restriction is that x cannot be -3 or -4, as either value will lead to 0 on the bottom. All other x values can be accepted.
Hope this helped.
