Ta có :
$\begin{array}{l}
{x^2} + 3x - 4 = {x^2} - x + 4x - 4\\
= x\left( {x - 1} \right) + 4\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x + 4} \right)\\
\Rightarrow {x^2} + 3x - 4 = 0 \Leftrightarrow \left( {x - 1} \right)\left( {x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 4
\end{array} \right.
\end{array}$
Vậy $A = \left\{ { - 4;1} \right\}.$
