Answer
$$x=\frac{3}{2}\text{ and } x=-1$$
Work Step by Step
Subtract $3$ from each side to obtain:
$$2x^2-x-3=0$$
Factor the trinomial to obtain:
$$(2x-3)(x+1)=0$$
Use the Zero-{Product Property by equating each factor to zero, then solve each equation to obtain:
\begin{align*}
2x-3&=0 &\text{or}& &x+1=0\\
2x&=3 &\text{or}& &x=-1\\
x&=\frac{3}{2} &\text{or}& &x=-1
\end{align*}
Check:
When $x=\frac{3}{2}$:
\begin{align*}
2\left(\frac{3}{2}\right)^2-\frac{3}{2}&\stackrel{?}=3\\\\
2\left(\frac{9}{4}\right)-\frac{3}{2}&\stackrel{?}=3\\\\
\frac{18}{4}-\frac{3}{2}&\stackrel{?}=3\\\\
\frac{9}{2}-\frac{3}{2}&\stackrel{?}=3\\\\
\frac{6}{2}&\stackrel{?}=3\\\\
3&\stackrel{\checkmark}=3
\end{align*}
When $x=-1$:
\begin{align*}
2(-1)^2-(-1)&\stackrel{?}=3\\\\
2(1)+1&\stackrel{?}=3\\\\
3&\stackrel{\checkmark}=3
\end{align*}
Therefore, the solutions to the given equation are $x=\frac{3}{2}\text{ and } x=-1$.
