$2{x}^{2}+3x=-2$
Divide both sides of the equation by $2$${x}^{2}+\frac{ 3 }{ 2 }x=-1$
To complete the square, the same value needs to be added to both sides${x}^{2}+\frac{ 3 }{ 2 }x+?=-1+?$
To complete the square ${x}^{2}+\frac{ 3 }{ 2 }x+\frac{ 9 }{ 16 }={\left( x+\frac{ 3 }{ 4 } \right)}^{2}$ add $\frac{ 9 }{ 16 }$ to the expression${x}^{2}+\frac{ 3 }{ 2 }x+\frac{ 9 }{ 16 }=-1+?$
Since $\frac{ 9 }{ 16 }$ was added to the left-hand side, also add $\frac{ 9 }{ 16 }$ to the right-hand side${x}^{2}+\frac{ 3 }{ 2 }x+\frac{ 9 }{ 16 }=-1+\frac{ 9 }{ 16 }$
Use ${a}^{2}+2ab+{b}^{2}={\left( a+b \right)}^{2}$ to factor the expression${\left( x+\frac{ 3 }{ 4 } \right)}^{2}=-1+\frac{ 9 }{ 16 }$
Calculate the sum${\left( x+\frac{ 3 }{ 4 } \right)}^{2}=-\frac{ 7 }{ 16 }$
Solve the equation for $x$$\begin{array} { l }x=-\frac{ 3 }{ 4 }-\frac{ \sqrt{ 7 } }{ 4 }i,\\x=-\frac{ 3 }{ 4 }+\frac{ \sqrt{ 7 } }{ 4 }i\end{array}$
The equation has $2$ solutions$\begin{array} { l }x_1=-\frac{ 3 }{ 4 }-\frac{ \sqrt{ 7 } }{ 4 }i,& x_2=-\frac{ 3 }{ 4 }+\frac{ \sqrt{ 7 } }{ 4 }i\end{array}$