Solve for p:
(p^2+3p-28)/(p-9)=0
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Solve for p:
(p^2+3p-28)/(p-9)=0
Hello, mathlete!
It's pretty straight forward . . . exactly where is your difficulty?
Quote:
Solve for $\displaystyle p\!:\;\;\frac{p^2+3p-28}{p-9}\:=\:0$
Providing $\displaystyle p \neq 9$, we can multiply by $\displaystyle (p-9)$
. . and we have: .$\displaystyle p^2 + 3p - 28 \:=\:0$
. . which factors: .$\displaystyle (p-4)(p+7)\:=\:0$
. . and has roots: .$\displaystyle p \;=\;4,\:-7$
