Solve for: (x+38) * (x+1)=0

Expression: $\left( x+38 \right) \times \left( x+1 \right)=0$

Simplify the expression

${x}^{2}+x+38x+38=0$

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

$\begin{array} { l }x_1=-38,& x_2=-1\end{array}$

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