Question

$(x - 2) { }^{2} + 1 = 2x - 3 \\ solve \: in \: quadrtic \: equation$
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Answer 1

Answer:

${x}^{2} + 4 - 4x + 1 = 2x - 3 \\ {x}^{2} - 4x - 2x = - 3 - 4 - 1 \\ {x}^{2} - 6x + 8=0$

Answer 2

Answer:

$\huge\mathcal\purple{Solution:-}$

${(x -2)}^{2} + 1 = 2x - 3 \\ \\ {x}^{2} + 4 - 4x + 1 = 2x - 3 \\ \\ {x}^{2} + 5 - 4x = 2x - 3 \\ \\ {x}^{2} + 5 + 3 = 2x + 4x \\ \\ {x}^{2} + 8 = 6x \\ \\Then \\\\ {x}^{2} - 6x + 8 = 0 \\ \\ {x}^{2} - 2x - 4x + 8x = 0 \\ \\ ({x}^{2} - 2x ) - (4x - 8x) = 0 \\ \\ x ( x - 2) - 4 ( x - 2 ) =0 \\ \\ (x - 2) (x - 4) = 0 \\ \\ x - 2 = 0 \: or \: x - 4 = 0 \\ \\ x = 2 \: or \: x = 4$