need some quick help

Hello, mathlete!

It's pretty straight forward . . . exactly where is your difficulty?

Quote:

Solve for $p\!:\;\;\frac{p^2+3p-28}{p-9}\:=\:0$

Providing $p \neq 9$ , we can multiply by $(p-9)$

. . and we have: . $p^2 + 3p - 28 \:=\:0$

. . which factors: . $(p-4)(p+7)\:=\:0$

. . and has roots: . $p \;=\;4,\:-7$