#### Answer

$-1$ and $3$

#### Work Step by Step

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 $3$