## Calculus with Applications (10th Edition)

$-1$ and $3$
Set the $RHS$ to 0 by subtracting 3 and 2x from both sides: $x^{2}=3+2x\qquad.../-3-2x$ $x^{2}-2x-3=0$ See example 2 in R-2 (factoring trinomials). We search for factors of $-3$ that add up to $-2$ .... $1\times(-3)=-3,\ 1$+$(-3)=-2$ so our equation becomes $(x+1)(x-3)=0$ We apply the zero product rule (if the product is zero, at least one of the factors is zero) $x=-1 \qquad...$(first factor =0) or $x=3 \qquad...$(second factor =0) solutions: $-1$ and $3$