$$(2a+1)^2+(1-a)^2 = 2(6a+1)$$
Answer
$$ \begin{matrix}a_1 = 0 & a_2 = 2 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (2a+1)^2+(1-a)^2 &= 2(6a+1)&& \text{simplify left and right hand side} \\[1 em]4a^2+4a+1+1-2a+a^2 &= 12a+2&& \\[1 em]5a^2+2a+2 &= 12a+2&& \text{move all terms to the left hand side } \\[1 em]5a^2+2a+2-12a-2 &= 0&& \text{simplify left side} \\[1 em]5a^2+2a+2-12a-2 &= 0&& \\[1 em]5a^2-10a &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 5x^{2}-10x = 0 } $, first we need to factor our $ x $.
$$ 5x^{2}-10x = x \left( 5x-10 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 5x-10 = 0$.
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