${x}^{2}+x+38x+38=0$
Collect like terms${x}^{2}+39x+38=0$
Move the constant to the right-hand side and change its sign${x}^{2}+39x=-38$
To complete the square, the same value needs to be added to both sides${x}^{2}+39x+?=-38+?$
To complete the square ${x}^{2}+39x+\frac{ 1521 }{ 4 }={\left( x+\frac{ 39 }{ 2 } \right)}^{2}$ add $\frac{ 1521 }{ 4 }$ to the expression${x}^{2}+39x+\frac{ 1521 }{ 4 }=-38+?$
Since $\frac{ 1521 }{ 4 }$ was added to the left-hand side, also add $\frac{ 1521 }{ 4 }$ to the right-hand side${x}^{2}+39x+\frac{ 1521 }{ 4 }=-38+\frac{ 1521 }{ 4 }$
Use ${a}^{2}+2ab+{b}^{2}={\left( a+b \right)}^{2}$ to factor the expression${\left( x+\frac{ 39 }{ 2 } \right)}^{2}=-38+\frac{ 1521 }{ 4 }$
Calculate the sum${\left( x+\frac{ 39 }{ 2 } \right)}^{2}=\frac{ 1369 }{ 4 }$
Solve the equation for $x$$\begin{array} { l }x=-38,\\x=-1\end{array}$
The equation has $2$ solutions$\begin{array} { l }x_1=-38,& x_2=-1\end{array}$