Question
$$(x+3)^2-2 = 7$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -6 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (x+3)^2-2 &= 7&& \text{simplify left side} \\[1 em]x^2+6x+9-2 &= 7&& \\[1 em]x^2+6x+7 &= 7&& \text{move all terms to the left hand side } \\[1 em]x^2+6x+7-7 &= 0&& \text{simplify left side} \\[1 em]x^2+6x+7-7 &= 0&& \\[1 em]x^2+6x &= 0&& \\[1 em] \end{aligned} $$
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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