Just "Calculate" not "Caculader" :) Wrong vocabulary sir :D

x(x + 3) - x - 3 = 0

\(\Leftrightarrow x^2+3x-x-3=0\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

So, \(S=\left\{-1;3\right\}\)

2x - x + 3x - 2x = 0 

\(\Leftrightarrow x+x=0\)

\(\Rightarrow x=-x\)

We just have one answer for this math problem.

That's \(x=0\)

So, \(S=\left\{0\right\}\)

We have : x(x + 3) - x - 3 = 0

=> x(x + 3) - (x + 3) = 0

=> (x - 1)(x + 3) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

2x - x + 3x - 2x = 0

=> -x + 3x = 0

=> 2x = 0

=> x = 0