Answer
Solution set = $\displaystyle \{\frac{-5-\sqrt{13}}{2},\frac{-5+\sqrt{13}}{2}\}.$
Work Step by Step
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$x^{2}+5x+3=0 \quad\rightarrow \left\{\begin{array}{l}
a=1\\
b=5\\
c=3
\end{array}\right.$
$x=\displaystyle \frac{-5\pm\sqrt{5^{2}-4(1)(3)}}{2(1)}=\frac{-5\pm\sqrt{25-12}}{2}=\frac{-5\pm\sqrt{13}}{2}$
$x= \displaystyle \frac{-5-\sqrt{13}}{2}\qquad$ or$ \displaystyle \qquad x=\frac{-5+\sqrt{13}}{2}$
Solution set = $\displaystyle \{\frac{-5-\sqrt{13}}{2},\frac{-5+\sqrt{13}}{2}\}.$
