Solve the equation $\displaystyle{{x}}^{{2}}+{4}{x}+{3}={0}\usepackage{cmbright}\Huge\$ . The first step is shown.

$\displaystyle{{x}}^{{2}}+{1}{x}+{3}{x}+{3}={0}\usepackage{cmbright}\Huge\$

One solution is $\displaystyle{x}=-{1}\usepackage{cmbright}\Huge\$ . What is the second solution?

The second solution is $\displaystyle{x}=\usepackage{cmbright}\Huge\$ _____.