Answer: [tex]x=3\\x=-2[/tex]
Step-by-step explanation:
1. You have the following expression:
[tex]\frac{x^{2}+8x+4}{x^{2}-x-6}[/tex]
3. The denominatoir cannot be zero, so, you must find its roots:
[tex]x^{2}-x-6=0[/tex]
Factorizing it, you obtain:
[tex](x-3)(x+2)=0\\x=3\\x=-2[/tex]
Then:
[tex]\frac{x^{2}+8x+4}{(x-3)(x+2)}[/tex]
4. Therefore, the values that are discontinuity of [tex]\frac{x^{2}+8x+4}{x^{2}-x-6}[/tex] are:
[tex]x=3\\x=-2[/tex]
