Question

$(2x + 1 {)} ^{ 2} - (4 - {2x)}^{2} = 0$
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Answer 1

$\displaystyle \tt (2x+1)^2-(4-2x)^2=0\\ \displaystyle \tt 4x^2+4x+1-(16-16x+4x^2)=0\\ \displaystyle \tt 4x^2+4x+1-16+16x-4x^2=0\\ \displaystyle \tt 20x-15=0\\ \displaystyle \tt 20x=15\\ \displaystyle \tt x=15\div20\\ \displaystyle \tt \bold{x=0,75}$