## Intermediate Algebra (6th Edition)

The solutions are $3$ and $2$.
Isolate $0$ on the right side. This can be done by moving the terms $9(p+2)-20$ from the right to the left side. Note that when a term is moved to the other side of an equation, its sign changes to its opposite. Thus, the equation is equivalent to: $(p+2)^2-9(p+2)+20=0$ Let $u=p+2$. Rewrite the equation above using $u$ to obtain: $u^2-9u+20=0$ Factor the trinomial to obtain: $(u-5)(u-4)=0$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&u-5=0 &\text{or} & u-4=0 \\&u=5 &\text{or} & u=4 \end{array} Since $u=p+2$, then \begin{array}{ccc} \\&u=5 &\text{or} & u=4 \\&p+2=5 &\text{or} & p+2=4 \\&p=5-2 &\text{or} & p=4-2 \\&p=3 &\text{or} & p=2 \end{array} Therefore, the solutions are $3$ and $2$.