# need some quick help

• Jan 18th 2008, 01:24 PM
mathlete
need some quick help
Solve for p:

(p^2+3p-28)/(p-9)=0
• Jan 18th 2008, 02:56 PM
Soroban
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: . $p \;=\;4,\:-7$