$
\begin{array}{l}
x^2 + 4x = 0 \\
(x + 2)^2 - 4 = 0 \\
(x + 2)^2 = 4 \\
x + 2= -2 \vee x + 2 = 2 \\
x = - 4 \vee x = 0 \\
\end{array}
$
$
\begin{array}{l}
x^2 + 7x + 12\frac{1}{4} = 0 \\
\left( {x + 3\frac{1}{2}} \right)^2=0 \\
x + 3\frac{1}{2} = 0 \\
x =-3\frac{1}{2} \\
\end{array}
$
$
\begin{array}{l}
x^2 - 4x = 0 \\
(x - 2)^2 - 4 = 0 \\
(x - 2)^2 = 4 \\
x - 2 =-2 \vee x - 2 = 2 \\
x = 0 \vee x = 4 \\
\end{array}
$
$
\begin{array}{l}
x^2 = x \\
x^2 - x = 0 \\
\left( {x - \frac{1}{2}} \right)^2 - \frac{1}{4} = 0 \\
\left( {x - \frac{1}{2}} \right)^2 = \frac{1}{4} \\
x - \frac{1}{2} = - \frac{1}{2} \vee x - \frac{1}{2} = \frac{1}{2} \\
x = 0 \vee x = 1 \\
\end{array}
$
$
\begin{array}{l}
x^2 + 8x + 4 = 0 \\
(x+4)^{2}-16+4=0 \\
(x+4)^{2}-12=0 \\
(x+4)^{2}=12 \\
x+4=-\sqrt{12}\vee x+4=\sqrt{12} \\
x=-4-\sqrt{12}\vee x=-4+\sqrt{12} \\
\end{array}
$
$
\begin{array}{l}
x^2 = 5x + 4 \\
x^2 - 5x = 4 \\
\left( {x - 2\frac{1}{2}} \right)^2 - 6\frac{1}{4} = 4 \\
\left( {x - 2\frac{1}{2}} \right)^2 = 10\frac{1}{4} \\
x - 2\frac{1}{2} = - \sqrt {10\frac{1}{4}} \vee x - 2\frac{1}{2} = \sqrt {10\frac{1}{4}} \\
x = 2\frac{1}{2} - \frac{1}{2}\sqrt {41} \vee x = 2\frac{1}{2} + \frac{1}{2}\sqrt {41} \\
\end{array}
$
$
\begin{array}{l}
x^2 + 2x = 31 \\
(x + 1)^2 - 1 = 31 \\
(x + 1)^2 = 32 \\
x + 1 = - \sqrt {32} \vee x + 1 = \sqrt {32} \\
x = - 1 - 4\sqrt {2} \vee x = - 1 + 4\sqrt {2} \\
\end{array}
$
$
\begin{array}{l}
2x^2 + 4x = 3 \\
2\left( {x^2 + 2x} \right) = 3 \\
2((x + 1)^2 - 1) = 3 \\
2(x + 1)^2 - 2 = 3 \\
2(x + 1)^2 = 5 \\
(x + 1)^2 = 2\frac{1}{2} \\
x + 1 = - \sqrt {2\frac{1}{2}} \vee x + 1 = \sqrt {2\frac{1}{2}} \\
x = - 1 - \frac{1}{2}\sqrt {10} \vee x = - 1 + \frac{1}{2}\sqrt {10} \\
\end{array}
$
$
\begin{array}{l}
x^2 - 27 = 2x \\
x^2 - 2x - 27 = 0 \\
(x - 1)^2 - 28 = 0 \\
x = 1 - 2\sqrt {7} \vee x = 1 + 2\sqrt {7} \\
\end{array}
$
$
\begin{array}{l}
(x + 1)(x + 5) = 2x + 13 \\
x^2 + 6x + 5 = 2x + 13 \\
x^2 + 4x + 5 = 13 \\
(x + 2)^2 + 1 = 13 \\
(x + 2)^2 = 12 \\
x = - 2 - 2\sqrt {3} \vee x = - 2 +2\sqrt {3} \\
\end{array}
$