Answer
$$\left\{-\dfrac{3}{2} - \dfrac{\sqrt{17}}{2}, -\dfrac{3}{2}+\dfrac{\sqrt{17}}{2}\right\}$$
Work Step by Step
Recall:
To complete the square of $x^2+bx$, add $\left(\dfrac{b}{2}\right)^2$.
Complete the square by adding $\left(\dfrac{3}{2}\right)^2$ to both sides of the equation to obtain:
\begin{align*}
x^2+3x+\left(\frac{3}{2}\right)^2&=2+\left(\frac{3}{2}\right)^2\\\\
x^2+3x+\frac{9}{4}&=2+\frac{9}{4}\\\\
\left(x+\frac{3}{2}\right)^2&=\frac{8}{4}+\frac{9}{4}\\\\
\left(x+\frac{3}{2}\right)^2&=\frac{17}{4}\\\\
\end{align*}
Take the square root of both sides:
\begin{align*}
\sqrt{\left(x+\frac{3}{2}\right)^2}&=\pm\sqrt{\frac{17}{4}}\\\\
x+\frac{3}{2}&=\pm\frac{\sqrt{17}}{2}\\\\
x&=-\frac{3}{2}\pm\frac{\sqrt{17}}{2}\\\\
\end{align*}
Thus, the solution set is:
$$\left\{-\dfrac{3}{2} - \dfrac{\sqrt{17}}{2}, -\dfrac{3}{2}+\dfrac{\sqrt{17}}{2}\right\}$$
