Question
$$(x-1)(x-5) = 5$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 6 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (x-1)(x-5) &= 5&& \text{simplify left side} \\[1 em]x^2-5x-x+5 &= 5&& \\[1 em]x^2-6x+5 &= 5&& \text{move all terms to the left hand side } \\[1 em]x^2-6x+5-5 &= 0&& \text{simplify left side} \\[1 em]x^2-6x+5-5 &= 0&& \\[1 em]x^2-6x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{2}-6x = 0 } $, first we need to factor our $ x $.
$$ x^{2}-6x = x \left( x-6 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ x-6 = 0$.
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